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Here are a few recommended readings before getting started with this lesson.
A quadratic equation is a polynomial equation of second degree, meaning the highest exponent of its monomials is 2. A quadratic equation can be written in standard form as follows.
ax2+bx+c=0
f(x)=2x2−2x−12 | |||
---|---|---|---|
Direction | Vertex | Axis of Symmetry | y-intercept |
a>0⇒ upward | (21,-225) | x=21 | (0,-12) |
In addition to these points, the reflection of the y-intercept across the axis of symmetry is at (1,-12), which is also on the parabola. Now, the graph can be drawn.
Now, find all the points on the graph that have a y-coordinate equal to 0.
The x-coordinates of the identified points solve the equation f(x)=0.
In the example, the x-coordinates of the points where the curve intercepts the x-axis are -2 and 3. Note that solving an equation graphically does not necessarily lead to an exact answer. To verify a solution, substitute it into f(x) and evaluate the expression. Verifying the solution x=-2 is done by evaluating f(-2).x=-2
Calculate power
Multiply
a(-b)=-a⋅b
a−(-b)=a+b
Add and subtract terms
Solution | Substitute | Evaluate |
---|---|---|
x=-2 | f(-2)=2(-2)2−2(-2)−12 | f(-2)=0 ✓ |
x=3 | f(3)=2(3)2−2(3)−12 | f(3)=0 ✓ |
For each quadratic equation, draw its related quadratic function to determine the number of solutions.
Paulina models the flight of a rocket she built using a quadratic function, where h is the height of the rocket in meters after t seconds.
f(t)=-16t2+112t−192 | |||
---|---|---|---|
Direction | Vertex | Axis of Symmetry | y-intercept |
a<0⇓downward
|
(3.5,4) | x=3.5 | (0,-192) |
In addition to the vertex and the y-intercept, the reflection of the y-intercept across the axis of symmetry, which is at (7,-192), is also on the parabola. With this information, the graph of the function can be drawn.
The parabola intersects the x-axis twice. The points of intersection are (3,0) and (4,0). Therefore, the equation -16t2+112t−192=0 has two solutions, t1=3 and t2=4.g(t)=-16t2+112t−196 | |||
---|---|---|---|
Direction | Vertex | Axis of Symmetry | y-intercept |
a<0⇓downward
|
(3.5,0) | x=3.5 | (0,-196) |
The reflection of the y-intercept across the axis of symmetry is at (7,-196), which is also on the parabola. Now, the graph can be drawn.
This time, there is only one x-intercept.
The point of intersection is (3.5,0). Therefore, the equation -16t2+112t−196=0 has one solution, t=3.5.Tadeo wants to form a square and a right triangle whose areas are equal. The side lengths of the geometric shapes are expressed in the diagram.
The area of the square is the square of 0.5x, and the area of the triangle is half the product of 2 and 0.5x+2.
Area of Square | Area of Triangle |
---|---|
(0.5x)2 | 21(2)(0.5x+2) |
0.25x2 | 0.5x+2 |
y=0.25x2−0.5x−2 | |||
---|---|---|---|
Direction | Vertex | Axis of Symmetry | y-intercept |
a>0⇓upward
|
(1,-2.25) | x=1 | (0,-2) |
The reflection of the y-intercept across the axis of symmetry (2,-2) is also on the parabola. With this information, the graph can be drawn.
From the graph, the x-intercepts of the function can be identified.
The parabola intersects the x-axis twice. The points of intersection are (-2,0) and (4,0). Therefore, the quadratic equation has two solutions, x1=-2 and x2=4.Area of the Square | Area of the Triangle |
---|---|
f(x)=(0.5x)2 | g(x)=21⋅2⋅(0.5x+2) |
f(x)=0.25x2 | g(x)=0.5x+2 |
To draw the graphs of these functions, a table of values will be made.
x | f(x)=0.25x2 | g(x)=0.5x+2 |
---|---|---|
-3 | 0.25(-3)2=2.25 | 0.5(-3)+2=0.5 |
0 | 0.25(0)2=0 | 0.5(0)+2=2 |
3 | 0.25(3)2=2.25 | 0.5(3)+2=3.5 |
(a−b)2=a2−2ab+b2
Distribute -0.5
Subtract term
LHS−4x=RHS−4x
LHS+0.5x2=RHS+0.5x2
y=-0.7x2+0.8x+8 | |||
---|---|---|---|
Direction | Vertex | Axis of Symmetry | y-intercept |
a<0⇓downward
|
(74,35288) | x=74 | (0,8) |
x=-3
Calculate power
Multiply
Add and subtract terms
The x-intercepts of the graph can now be identified.
The parabola intersects the x-axis twice. From the graph only one of the intercepts is identified easily, which is (4,0). The other intercept is between -3 and -2. However, this intercept can be ignored because x represents horizontal distance and distance cannot be negative. Therefore, in this context, only x=4 makes sense.