2. Graphing Quadratic Functions Using Standard Form
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Chapter 8
2.
Graphing Quadratic Functions Using Standard Form
This lesson delves into the intricacies of graphing quadratic functions, particularly when they are presented in standard form. It emphasizes the importance of understanding the vertex, axis of symmetry, and other key characteristics of a parabola. Through various examples, the lesson illustrates how to determine whether a quadratic function is expressed in standard form and how to graph it accordingly. The use cases highlighted include real-world scenarios like modeling the trajectory of a kicked ball and understanding the height of fireworks after launch. These practical applications underscore the relevance of mastering quadratic functions in everyday life.
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Lesson Settings & Tools
| | 13 Theory slides |
| | 10 Exercises - Grade E - A |
| | Each lesson is meant to take 1-2 classroom sessions |
Graphing Quadratic Functions Using Standard Form
Slide of 13
This lesson explains what the standard form of a quadratic function is and how to draw its graph. How to determine various characteristics of a quadratic function such as the axis of symmetry, the vertex, and the y-intercept will then be understood.
Catch-Up and Review
Here are a few recommended readings before getting started with this lesson.
Explore
Expanding the Vertex Form of a Quadratic Function
Consider the vertex form of a quadratic function y=a(x−p)2+q and its corresponding parabola. In the applet, adjust the parameters of each parabola's vertex. Investigate how the expanded form of the equation is rewritten according to the new parameters. 
Since the obtained function is equivalent to the given function in vertex form, the parabola also corresponds to the obtained function.
Explore
Expanding the Intercept Form of a Quadratic Function
Consider the intercept form of a quadratic function and its corresponding parabola. In the applet, adjust the parameters of each parabola's x-intercepts. Investigate how the expanded form of the equation is rewritten according to the new parameters. 
Since the obtained function is equivalent to the given function in intercept form, the parabola also corresponds to the obtained function.
Discussion
Standard Form of a Quadratic Function
Besides the intercept and the vertex forms, another essential and common form of a quadratic function is its standard form.
The standard form of a quadratic function is a quadratic function written in a specific format.
f(x)=ax2+bx+c
Here, a, b, and c are real numbers with a=0. The term with the highest degree — the quadratic term — is written first, then the linear term, followed by the constant term. The standard form of the function can be used to determine the direction of the parabola, the y-intercept, the axis of symmetry, and the vertex.
| Direction of the Graph | Opens upward when a>0 |
|---|---|
| Opens downward when a<0 | |
| y-intercept | c |
| Axis of Symmetry | x=-2ab |
| Vertex | (-2ab,f(-2ab)) |
Pop Quiz
Is the Quadratic Function Written in Standard Form?
Determine whether the given quadratic function is expressed in standard form.
Discussion
Graphing a Quadratic Function in Standard Form
Given a quadratic function in standard form, some characteristics of its corresponding parabola can be determined. Consider an example quadratic function.
f(x)=x2−4x+3
To draw the graph of the function written in standard form, there are five steps to follow.
1
Identify and Graph the Axis of Symmetry
expand_more The axis of symmetry can be found by determining a and b in the form ax2+bx+c.
The axis of symmetry of the function is the vertical line with equation x=2. 
f(x)=x2−4x+3⇕f(x)=1x2+(-4)x+3
In the given function, a and b are 1 and -4, respectively. Now, these values will be substituted into the formula for the equation of the axis of symmetry.
x=-2ab
SubstituteII
a=1, b=-4
x=-2(1)-4
▼
Evaluate right-hand side
IdPropMult
Identity Property of Multiplication
x=-2-4
RemoveNegFracAndNum
-b-a=ba
x=24
CalcQuot
Calculate quotient
x=2
2
Determine and Plot the Vertex
expand_more The axis of symmetry intersects the parabola at its vertex. Therefore, x=2 is the x-coordinate of the vertex. To find the y-coordinate of the vertex, x=2 can be substituted into the function rule.
The vertex of the parabola is at (2,-1). 
3
Determine and Plot the y-intercept
expand_more The y-intercept can be determined by using the constant term c of the given function. In this case, c is equal to 3. This means that the y-intercept occurs at (0,3).
4
Reflect the y-intercept Across the Axis of Symmetry
expand_more The axis of symmetry divides the graph into two mirror images. Therefore, the reflection of the y-intercept across the axis of symmetry is also on the parabola.
If the vertex of a quadratic function lies on the y-axis, then any point that lies on the graph other than the vertex should be found and reflected across the axis of symmetry. In this case, the y-axis is the axis of symmetry.
5
Draw the Parabola
expand_more Finally, connect the points with a smooth curve to graph the parabola.
The graph of the function opens upward. This is expected, since the value of a is 1, which is a positive number.
Pop Quiz
Determining the Vertex of a Quadratic Function Written in Standard Form
Find the coordinates of the vertex of the parabola that corresponds to the quadratic function written in standard form. If necessary, round the answer to 2 decimal places.
Example
Will a German Shepherd Jump Over the Fence?
Zain lives in a quiet rural town where it seems not a lot happens. He decides to adopt a German shepherd puppy. After a few short weeks the puppy is growing so fast, they are afraid that soon the pup will be able to jump over the fence around their garden!
External credits: @brgfx
h(x)=-0.075x2+1.35x
Here, x is the dog's horizontal distance from the jump spot and h(x) is the height of that jump. Both values are given in feet. a Draw the function's graph.
b If the height of the fence is 7 feet, will their dog, when it is an adult, manage to jump over the fence?
c Zosia and Zain made one more observation about their puppy's mad hops. The puppy can already jump, horizontally, over a 5-foot-wide hole. How much further will the dog be able to jump as an adult?
Answer
a
b No
c 13 feet
Hint
a Start by determining the axis of symmetry and the vertex of the parabola. Then, find two more points that lie on the curve.
b Consider the coordinates of the parabola's vertex.
c Use the graph from Part A.
Solution
a Consider the given quadratic function.
h(x)=-0.075x2+1.35x
This function is written in standard form. To graph a quadratic function written in this form, there are five steps to follow. - Identify and graph the axis of symmetry
- Determine and plot the vertex
- Determine and plot the y-intercept
- Reflect the y-intercept in the axis of symmetry
- Draw the parabola
These steps will be done one at a time.
The Axis of Symmetry
To find the equation of the axis of symmetry, the coefficients a, b, and c should first be found. Consider the given function.h(x)=-0.075x2+1.35x⇕h(x)=-0.075x2+1.35x+0
Here, a=-0.075, b=1.35, and c=0. The axis of symmetry is a vertical line with equation x=2a-b.
The axis of symmetry is the vertical line x=9. 
The Vertex
The vertex of the function lies on the axis of symmetry. This means that the vertex's x-coordinate is 9. Now, to find the y-coordinate, x=9 will be substituted into the function rule.h(x)=-0.075x2+1.35x
Substitute
x=9
h(9)=-0.075(9)2+1.35(9)
▼
Evaluate right-hand side
CalcPowProd
Calculate power and product
h(9)=-0.075(81)+12.15
MultNegPos
(-a)b=-ab
h(9)=-6.075+12.15
AddTerms
Add terms
h(9)=6.075
The y-intercept
The y-intercept is given by the constant term c of the function rule. In this case this term is equal to 0, which means that the y-intercept occurs at (0,0) — the origin.
Reflecting the y-intercept
Now, another point that lies on the parabola can be found by reflecting the y-intercept in the axis of symmetry.
The third point that lies on the parabola is at (18,0).
Drawing the Parabola
Finally, the points will be connected with a smooth curve to graph the parabolic shape. Since the function represents a dog's jump, negative values of the function will not be included.
b To determine whether the adult dog will be able to jump over the fence, the fence will be drawn in the same coordinate plane. Assume that the fence is placed at the axis of symmetry, where the height of the jump would the greatest.
c It is known that the dog can jump over a hole whose width is 5 feet. Consider the graph of the function once again and note the horizontal distance that the dog will be able to jump as an adult.
18−5=13 feet
What a pup!
Discussion
Rewriting a Quadratic Function in Standard Form
Depending on the conditions, it is convenient to rewrite a quadratic function given in intercept or vertex form in its standard form.
Both the vertex and intercept forms of a quadratic function can always be rewritten in standard form.
| Form | Equation | How to Rewrite? |
|---|---|---|
| Vertex Form | y=a(x−h)2+k | Expand (x−h)2, distribute a, and combine like terms. |
| Intercept Form (also called Factored Form) |
y=a(x−p)(x−q) | Multiply a(x−p)(x−q) and combine like terms. |
Example
Graphing Quadratic Functions to Represent Profit
The mayor of Zain's sleepy town wants to do something exciting for the community. Lucky her, she noticed an increasing interest in people taking up skateboarding, so she knew that building a new skate ramp would be a huge hit.
She was right! Now a local skateboard company named SuTeKi Sk8 sees this as an opportunity to make some big sales. SuTeKi Sk8 investigates the profit of its two coolest models, Sketchtastic and Sugoi Sketch. They collect data and find a relationship between the price and the profit on sales of both skateboards. The following quadratic functions represent their results.
| Skateboard | Relationship |
|---|---|
| Sketchtastic | y=-0.015(x−50)2+37 |
| Sugoi Sketch | y=-0.008(x−65)2+28 |
In these function rules, x is the price in dollars and y is the corresponding profit in thousands of dollars.
a Rewrite the quadratic functions in standard form and graph them in the same coordinate plane.
c Using the graph, estimate the profit on sales if they set the price at $80 for each model. Then, compare the results with the exact values.
Answer
a Standard Forms:
| Skateboard | Vertex Form | Standard Form |
|---|---|---|
| Sketchtastic | y=-0.015(x−50)2+37 | y=-0.015x2+1.5x−0.5 |
| Sugoi Sketch | y=-0.008(x−65)2+28 | y=-0.008x2+1.04x−5.8 |
Graphs:
b See solution.
c Estimation:
| Skateboard | Profit on Sales |
|---|---|
| Sketchtastic | ≈$23000 |
| Sugoi Sketch | ≈$26000 |
Comparison: See solution.
Hint
a To rewrite the quadratic functions in standard form, start by expanding the squared expression. To draw a parabola, determine the axis of symmetry and the vertex. Then, find two more points on the curve.
b The vertex of a parabola represents the maximum or minimum value of the function.
c To estimate the profit, look at the values of the function at x=80. Then, substitute x=80 into the function rules to calculate the exact values.
Solution
a The quadratic functions that represent the data about the skateboard models are given in vertex form.
| Vertex Form: y=a(x−p)2+q | |
|---|---|
| Sketchtastic | Sugoi Sketch |
| y=-0.015(x−50)2+37 | y=-0.008(x−65)2+28 |
y=-0.015(x−50)2+37
▼
Rewrite
ExpandNegPerfectSquare
(a−b)2=a2−2ab+b2
y=-0.015(x2−2x(50)+502)+37
CalcPowProd
Calculate power and product
y=-0.015(x2−100x+2500)+37
Distr
Distribute -0.015
y=-0.015x2+1.5x−37.5+37
AddTerms
Add terms
y=-0.015x2+1.5x−0.5
| Sketchtastic | Sugoi Sketch | |
|---|---|---|
| Given Function | y=-0.015(x−50)2+37 | y=-0.008(x−65)2+28 |
| Standard Form | y=-0.015x2+1.5x−0.5 | y=-0.008x2+1.04x−5.8 |
The functions can now be graphed using the standard form. To do so, there are five steps to follow.
- Identify and graph the axis of symmetry.
- Determine and plot the vertex.
- Determine and plot the y-intercept.
- Reflect the y-intercept in the axis of symmetry.
- Draw the parabola.
The axis of symmetry is given by the equation x=2a-b, where a is the coefficient of the x2-term and b is the linear coefficient. Use a calculator to evaluate the equations if necessary.
| Axis of Symmetry: x=2a-b | |
|---|---|
| y=-0.015x2+1.5x−0.5 | y=-0.008x2+1.04x−5.8 |
| x=2(-0.015)-1.5 | x=2(-0.008)-1.04 |
| x=50 | x=65 |
The axes of symmetry of the parabolas are the vertical lines with equations x=50 and x=65.
The vertices of the parabolas lie on the corresponding axes of symmetry. This means that x=50 and x=65 are the x-coordinates of the vertices. Now, these values will be substituted for x into the function rules to find the y-coordinates.
| y=-0.015x2+1.5x−0.5 | y=-0.008x2+1.04x−5.8 | |
|---|---|---|
| x-coordinate | 50 | 65 |
| y-coordinate | y=-0.015(50)2+1.5(50)−0.5 ⇕ y=37 |
y=-0.008(65)2+1.04(65)−5.8 ⇕ y=28 |
| Vertex | (50,37) | (65,28) |
The vertices can now be added to the graph.
The y-intercepts are given by the constant term of the function rules.
| Function Rule | y-intercept |
|---|---|
| y=-0.015x2+1.5x−0.5 | -0.5 |
| y=-0.008x2+1.04x−5.8 | -5.8 |
Next, another point that lies on each parabola can be found by reflecting the y-intercept in the axis of symmetry.
Finally, the corresponding points will be connected with a smooth curve to graph the parabolic shapes.
b In the given functions, x is the price in dollars and y is the corresponding the profit on sales in thousands of dollars. Both functions reach their maximum values at their corresponding vertices.
The x-coordinate of each vertex is the price of the corresponding skateboard at which the profit of the company is maximized. This means that the vertex represents the optimal price and its corresponding profit. The company should use these values to make the highest return on sales.
c To estimate the profit of the skateboards if the price is set at $80, consider the graph of the functions once again. Since that price is given, a line corresponding to the price of $80 will be added to the graph.
From the graph, it can be seen that the profits of skateboards Sketchtastic and Sugoi Sketch are about 23 and 26 dollars in the thousands, respectively. To verify this, x=80 will be substituted in both function rules.
| Sketchtastic | Sugoi Sketch | |
|---|---|---|
| x | 80 | 80 |
| Substitute | -0.015(80)2+1.5(80)−0.5 | -0.008(80)2+1.04(80)−5.8 |
| Evaluate | 23.5 | 26.2 |
The estimated values are slightly less than the exact values.
Example
Using a Quadratic Function to Model the Trajectory of a Kicked Ball
Zain's sleepy town is becoming more exciting these days. Two rival soccer teams, feeling the town's energy, meet at the park. The two captains begin a duel of sorts, and a soccer ball is kicked from the ground level towards a goal 80 feet away. The ball follows a parabolic path and is caught at the height of 487 feet by a goalkeeper. The goalkeeper is standing 15 feet in front of the goal.
There are two x-intercepts. These are the starting point at (0,0) and the goal's line at (80,0). Therefore, to write the quadratic function representing the path of the ball, the intercept form can be used. The zeros of the parabola are 0 and 80.
In the given scenario, the ball does not hit the goal's line. Instead, the goalkeeper, standing 15 feet in front of the goal, catches the ball at a height of 487 feet.
When the goalkeeper catches the ball, its horizontal distance from the starting point is 65 feet. This means that the value of the function at x=65 is y=487. These values can be substituted into the intercept form to find the missing coefficient a.
Finally, the quadratic function that represents the ball's path can be written.
The parabola's vertex lies on the axis of symmetry. This means that x=40 is the x-coordinate of the vertex. Now, this value will be substituted into the function rule to find the y-coordinate.
The vertex of the parabola is at (40,8). 
External credits: Jeffrey F Lin
a In its intercept form, write the equation of the quadratic function that represents the path of the ball, assuming that it would hit the goal's line.
b Rewrite the obtained function in standard form. Then, graph the function.
c What is the maximum height of the ball?
d Is the ball already falling when its horizontal distance from the kicker is 50 feet?
Answer
a y=-0.005x(x−80)
b Standard Form: y=-0.005x2+0.4x
Graph:
c 8 feet
d Yes, see solution.
Hint
a Illustrate the given situation on a coordinate plane. Mark the starting position of the ball and the goal's line on the x-axis. Then, use intercept form of a quadratic function to find the function rule.
b Rewrite the function obtained in Part A by expanding the function rule. Next, determine the axis of symmetry and the vertex of the given function. Then, find two more points that lie on the parabola.
c At what point does a parabola reach its maximum or minimum value?
d Consider the graph of the function. For what values of x is the function decreasing?
Solution
a To write the quadratic function that represents the path of the ball, consider all the information given.
- A player kicks the ball from a distance of 80 feet.
- If the goalkeeper does not catch the ball, it will hit the goal's line.
- The goalkeeper — standing 15 feet in front of the goal — catches the ball at a height of 487 feet.
y=ax(x−80)
SubstituteII
x=65, y=487
487=a(65)(65−80)
▼
Solve for a
SubTerms
Subtract terms
487=a(65)(-15)
MultPosNeg
a(-b)=-a⋅b
487=a(-975)
MixedToFrac
acb=ca⋅c+b
839=a(-975)
DivEqn
LHS/(-975)=RHS/(-975)
-97539/8=a
MoveNegDenomToFrac
Put minus sign in front of fraction
-97539/8=a
DivFracD
ba/c=b⋅ca
-780039=a
UseCalc
Use a calculator
-0.005=a
RearrangeEqn
Rearrange equation
a=-0.005
y=ax(x−80)⇓y=-0.005x(x−80)
b In Part A, a quadratic function written in intercept form has been obtained.
y=-0.005x(x−80)
To rewrite this function in standard form, the expression -0.005x will be distributed.
y=-0.005x(x−80)⇕y=-0.005x2+0.4x
The function can now be graphed using the standard form. To do so, there are five steps to follow. - Identify and graph the axis of symmetry
- Determine and plot the vertex
- Determine and plot the y-intercept
- Reflect the y-intercept in the axis of symmetry
- Draw the parabola
The y-intercept is given by the constant term of the function rule, which in this case is 0. Therefore, the y-intercept is at the origin (0,0).
Now, another point that lies on the parabola can be found by reflecting the y-intercept in the axis of symmetry.
The third point lies at (80,0). Finally, the three points will be connected with a smooth curve to graph the parabolic shape. Since the function represents height of the ball, negative values of the function will not be considered.
c To determine the maximum height of the ball, the maximum of the given function will be considered.
A parabola that opens downward reaches its maximum at the vertex. In Part B, it was obtained that the vertex of the given function is at (40,8). Therefore, the maximum height of the ball is 8 feet.
d Consider the graph of the given function once again. The horizontal distance d=50 from the kicker — who is placed at the origin — will be marked in the graph.
Recall that the x-coordinate of the vertex is 40. Since the given distance is greater than 40, it can be said that the ball is already falling at x=50.
If the goalie understands this math concept, and can apply it in the heat of a soccer match, they are surely going to be an amazing player and maybe some day a sports scientist. What exciting times in what Zain thought was a sleepy town!
Pop Quiz
Rewriting a Quadratic Function in Standard Form
The quadratic functions are given in either intercept or vertex form. State the coefficients a, b, and c of the standard form y=ax2+bx+c.
Closure
Graphing Quadratic Functions in Different Forms
The three most common forms of a quadratic function are standard form, intercept form, and vertex form.
| Form | Formula |
|---|---|
| Standard | y=ax2+bx+c |
| Intercept | y=a(x−p)(x−q) |
| Vertex | y=a(x−h)2+k |
In these forms, a is not equal to 0. Sometimes, the function rule of a quadratic function is not known. Instead of the function rule, the y-intercept and at least two more points that lie on the parabola may be given. In this case, the function can also be graphed using its standard form. The coefficients of the standard form can be found by following four steps.
- Use the y-intercept to find the constant term c
- Substitute the coordinates of the points into the general standard form for x and y
- Solve the obtained system of equations for a and b
- Write the standard form of the function
An example will be considered.
Example
Consider the y-intercept of a parabola and two of its points.y-intercept:Points: (0,1) (1,1),(2,5)
Now, the standard form of the function y=ax2+bx+c will be found by following the mentioned steps.
1
Use the y-intercept to Find c
expand_more The y-intercept occurs at (0,1). This means that c is equal to 1.
y=ax2+bx+c⇓y=ax2+bx+1
2
Substitute the Coordinates of the Points
expand_more Next, the coordinates of the given points will be substituted for x and y into the obtained partial equation. Each point will be considered one at a time starting with (1,1).
Similarly, the coordinates of the second point (2,5) will be substituted.
The obtained equations are a+b=0 and 4a+2b=4.
y=ax2+bx+1
SubstituteII
x=1, y=1
1=a(12)+b(1)+1
▼
Simplify
BaseOne
1a=1
1=a(1)+b(1)+1
IdPropMult
Identity Property of Multiplication
1=a+b+1
SubEqn
LHS−1=RHS−1
0=a+b
RearrangeEqn
Rearrange equation
a+b=0
y=ax2+bx+1
SubstituteII
x=2, y=5
5=a(22)+b(2)+1
▼
Simplify
CalcPow
Calculate power
5=a(4)+b(2)+1
CommutativePropMult
Commutative Property of Multiplication
5=4a+2b+1
SubEqn
LHS−1=RHS−1
4=4a+2b
RearrangeEqn
Rearrange equation
4a+2b=4
3
Solve the System of Equations
expand_more Now, consider the system of equations formed by the obtained equations.
{a+b=04a+2b=4(I)(II)
The system can be solved by using the Substitution Method. For simplicity, in this case a-variable will be isolated in Equation (I).
Now, Equation (II) will be solved for b.
{a=-b4(-b)+2b=4
▼
(II): Solve for b
MultPosNeg
(II): a(-b)=-a⋅b
{a=-b-4b+2b=4
AddTerms
(II): Add terms
{a=-b-2b=4
DivEqn
(II): LHS/(-2)=RHS/(-2)
⎩⎪⎨⎪⎧a=-bb=-24
MoveNegDenomToFrac
(II): Put minus sign in front of fraction
⎩⎪⎨⎪⎧a=-bb=-24
CalcQuot
Calculate quotient
{a=-bb=-2
Substitute
(I): b=-2
{a=-(-2)b=-2
NegNeg
(I): -(-a)=a
{a=2b=-2
4
Write the Standard Form
expand_more Finally, the standard form can be written.
y=2x2+(-2)x+1⇕y=2x2−2x+1
Graphing Quadratic Functions Using Standard Form
Exercises
Consider the following parabola.
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The vertex of the parabola is the highest or lowest point on the curve.
The parabola opens upward and therefore the vertex is its minimum point. We see the vertex is the point (2,-2).
Now we will find the equation of the axis of symmetry. The axis of symmetry is the vertical line passing through the vertex, and it divides the parabola into two mirrored images.
The equation of the axis of symmetry is x=2.
Finally, we can find the y-intercept. Recall that the y-intercept is the point where the graph intercepts the y-axis.
As we can see, the y-intercept is located at (0,1).
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Consider each of the quadratic functions.
f(x)=4x2−16x
Find the vertex of the graph of the function.
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g(x)=-6x2+3x
Determine the axis of symmetry of the given parabola.
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y=-x2+6x−2
What are the vertex and the axis of symmetry of the graph of the given function?
A.B.C.D.Vertex: (6,-2)Axis of Symmetry: x=6Vertex: (-3,-29)Axis of Symmetry: x=-3Vertex: (3,-2)Axis of Symmetry: x=3Vertex: (3,7)Axis of Symmetry: x=3
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We have a quadratic function written in standard form. f(x)= ax^2+ bx+ c This form can give us a lot of information about the parabola by observing the values of a, b, and c. f(x)=4x^2-16x ⇕ f(x) = 4x^2 + ( -16)x + 0 We see that for the given function a= 4, b= -16, and c= 0. Let's consider the point at which the curve of the parabola changes direction.
This point is the vertex of the parabola. If we want to calculate the x-coordinate of this point, we can substitute the given values of a and b into the expression - b2a and simplify.
Let's now substitute this value into the given function rule to find the y-coordinate of the vertex.
Therefore, the vertex is (2,- 16).
Similar to Part A, consider the given function written in standard form. g(x)=-6x^2+3x ⇕ g(x) = -6x^2 + 3x + 0 We see that for the given equation a= -6, b= 3, and c= 0. The x-coordinate of the vertex of the parabola defines the axis of symmetry. Once again, let's substitute the values of a and b into the expression - b2a and simplify.
Remember that the axis of symmetry is the vertical line through the vertex, and divides a parabola into two mirror images. Since every point on this line will have the same x-coordinate as the vertex, we can write its equation. x=1/4 ⇔ x = 0.25
Just as in Parts A and B, determine the values of a, b, and c in the given standard form of the quadratic function. y=- x^2+6x-2 ⇕ y = -1x^2 + 6x + ( -2) We see that for the given equation a= -1, b= 6, and c= -2. Now, let's find the x-coordinate of the vertex by substituting these values into the expression - b2a and simplify.
The x-coordinate of the vertex is 3. This means that we can write the equation of the axis of symmetry. x=3 Finally, let's substitute x=3 into the given function to find the y-coordinate of the vertex.
The vertex is (3,7). This, along the axis of symmetry we found, corresponds to option D.
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Consider each of the given quadratic functions.
y=3x2−12x+12
Determine whether the function has a minimum or a maximum value.
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y=-4x2−4x+4
Find the maximum value of the function.
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y=-2x2+4x+1
Determine whether the function has a minimum or a maximum value. Then, find this value and choose the correct option.
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We have a quadratic function written in standard form. y= ax^2+ bx+ c This equation can give us a lot of information about the parabola by observing the values of a, b, and c. Consider the given quadratic function. y = 3x^2 - 12x + 12 ⇕ y = 3x^2+( -12)x + 12 We see that for the given function rule a= 3, b= -12, and c= 12. The point at which the graph of a parabola changes direction also defines the maximum or minimum point of the graph. Whether the parabola has a minimum or maximum is determined by the value of a.
Since the given value of a is positive, the parabola reaches a minimum value at the vertex.
Similar to Part A, we are given a quadratic function written in standard form.
y = -4x^2 - 4x + 4
⇕
y = -4x^2+( -4)x + 4
We see that for the given function rule a= -4, b= -4, and c= 4. It is already given that the function reaches a maximum value. To find this value, we first need to determine the vertex of the function.
x-coordinate of the Vertex
We can find the x-coordinate of the vertex by substituting the values of a and b into the expression - b2a and simplifying.
y-coordinate of the Vertex
To find the y-coordinate of the vertex, think of y as a function of x. By substituting the x-coordinate of the vertex into the given equation and simplifying, we will get the y-coordinate of the vertex.
This means that y=5 is the maximum value of the given function.
Just as in Part A and Part B, consider the given quadratic function.
y = -2x^2 + 4x + 1
We see that for the given function rule a= -2, b= 4, and c= 1.
The x-coordinate of the Vertex
Consider the point at which the curve of the parabola changes direction.
This point is the vertex of the parabola. Once again, if we want to calculate the x-coordinate of this point, we can substitute the given values of a and b into the expression - b2a and simplify.
The y-coordinate of the Vertex
The point at which the graph of a parabola changes direction also defines the maximum or minimum point of the graph. Whether the parabola has a minimum or maximum is determined by the value of a.
Since the given value of a is negative, the parabola has a maximum value at the vertex. To find this value, think of y as a function of x. By substituting the x-coordinate of the vertex into the given equation and simplifying, we will obtain the y-coordinate of the vertex.
Therefore, the maximum value of the function is 3.
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A fireworks stand in the local neighborhood introduced a new firework model that reaches great heights after it is launched.
h=-32t2+384t
In this function, h represents the height of the firework — in feet — t seconds after it is launched. We assume that the firework explodes at its highest point.{"type":"text","form":{"type":"math","options":{"comparison":"1","nofractofloat":false,"keypad":{"simple":true,"useShortLog":false,"variables":[],"constants":[]},"rendered":false},"text":"<span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><\/span><\/span>"},"formTextBefore":"<span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><span class=\"base\"><span class=\"strut\" style=\"height:0.61508em;vertical-align:0em;\"><\/span><span class=\"mord mathdefault\">t<\/span><span class=\"mspace\" style=\"margin-right:0.2777777777777778em;\"><\/span><span class=\"mrel\">=<\/span><\/span><\/span><\/span>","formTextAfter":"seconds","answer":{"text":["6"]}}
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We have a quadratic function written in standard form. h= at^2 + bt + c This equation can give us information about the parabola by observing the values of a, b, and c. h = -32t^2+384t ⇕ h = -32t^2+ 384t+ 0 We see that for the given equation a= -32, b= 384, and c= 0. Since -16<0, the parabola opens downward and the vertex is a maximum point. Therefore, the x-coordinate of the vertex will give us the time when the firework explodes. To calculate it, we can substitute the given values of a and b into the expression - b2a and simplify.
Therefore, the firework will explode 6 seconds after we launch it.
In Part A we have found that the firework reaches its maximum height for t=6. Also, at this time is when it explodes. By substituting this value into the given function rule, we can find its maximum height.
The firework will explode at 1152 feet. What a height!
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Consider a parabola with an unknown equation.
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We know that the x-coordinate of the vertex defines the axis of symmetry of the parabola.
| Vertex | Axis of Symmetry |
|---|---|
| ( 2,5) | x= 2 |
Therefore, the axis of symmetry is the vertical line x= 2. The point ( -1,-3) is 3 units to the left from the axis of symmetry. On the other side of the axis of symmetry, there is a point that lies on the parabola which is also 3 units away.
Note that the x-coordinate of this point is 5. This means that ( 5, -3) is another point on the parabola.
We are given two points that lie on the parabola, (0, 4) and (4, 4). Notice that the y-coordinates of the points are both equal to 4.
Recall that the axis of symmetry is a vertical line that divides the parabola in two mirror images. This means that it should lie midway between the points. Consequently, (0,4) and (4,4) lie on the opposite sides of the axis of symmetry. Let's mark the distance d from each point to the axis in the diagram.
From the diagram, we can see that the distance between the points is equal to 2d. Because the y-coordinates of the points are equal, this distance is the difference between the x-coordinates of the points. 2d = 4 - 0 ⇕ d = 2 Now, the equation of the axis of symmetry can by found by adding d=2 to the x-coordinate of the first point.
Since the vertex of the parabola lies on the axis of symmetry, the obtained equation gives us that the x-coordinate of the vertex is 2. Finally, let's graph an example parabola that satisfies the given conditions.
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y=ax2+bx+c
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Let's start by recalling the equation of the axis of symmetry of the graph of a quadratic function. ccc Quadratic Function & & Axis of Symmetry [0.8em] y=ax^2+bx+c & & x=- b/2a In the case of b=0, we can find the equation of the axis of symmetry by substituting 0 for b into the formula x=- b2a. Because the expression 2a is in the denominator of the equation, we need to assume that a is not equal to 0.
When b=0, the axis of symmetry is the vertical line x=0.
Once again, let's recall the equation of the axis of symmetry.
x =- b/2a
Notice that the value of c is not included in the equation. This means that the axis of symmetry depends only on the values of a and b. Therefore, c does not influence the equation of the axis of symmetry.
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Graphing Quadratic Functions Using Standard Form
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