Unlocking the Secrets of Solving Quadratic Equations Graphically

$f (x) = 2 x^{2} - 2 x - 12$
Direction	Vertex	Axis of Symmetry	$y -$ intercept
$a > 0 \Rightarrow$ upward	$(\frac{1}{2}, - \frac{2 5}{2})$	$x = \frac{1}{2}$	$(0, - 12)$

f (x) = 2 x^{2} - 2 x - 12

Direction

Vertex

Axis of Symmetry

y -

intercept

a > 0 \Rightarrow

upward

(\frac{1}{2}, - \frac{2 5}{2})

x = \frac{1}{2}

(0, - 12)

Solution	Substitute	Evaluate
$x = - 2$	$f (- 2) = 2 (- 2)^{2} - 2 (- 2) - 12$	$f (- 2) = 0 ✓$
$x = 3$	$f (3) = 2 (3)^{2} - 2 (3) - 12$	$f (3) = 0 ✓$

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 ✓

$f (t) = - 16 t^{2} + 112 t - 192$
Direction	Vertex	Axis of Symmetry	$y -$ intercept
$a < 0 ⇓ downward$	$(3.5, 4)$	$x = 3.5$	$(0, - 192)$

f (t) = - 16 t^{2} + 112 t - 192

Direction

Vertex

Axis of Symmetry

y -

intercept

a < 0 ⇓ downward

(3.5, 4)

x = 3.5

(0, - 192)

$g (t) = - 16 t^{2} + 112 t - 196$
Direction	Vertex	Axis of Symmetry	$y -$ intercept
$a < 0 ⇓ downward$	$(3.5, 0)$	$x = 3.5$	$(0, - 196)$

g (t) = - 16 t^{2} + 112 t - 196

Direction

Vertex

Axis of Symmetry

y -

intercept

a < 0 ⇓ downward

(3.5, 0)

x = 3.5

(0, - 196)

Area of Square	Area of Triangle
$(0.5 x)^{2}$	$\frac{1}{2} (2) (0.5 x + 2)$
$0.25 x^{2}$	$0.5 x + 2$

Area of Square

Area of Triangle

(0.5 x)^{2}

\frac{1}{2} (2) (0.5 x + 2)

0.25 x^{2}

0.5 x + 2

$y = 0.25 x^{2} - 0.5 x - 2$
Direction	Vertex	Axis of Symmetry	$y -$ intercept
$a > 0 ⇓ upward$	$(1, - 2.25)$	$x = 1$	$(0, - 2)$

y = 0.25 x^{2} - 0.5 x - 2

Direction

Vertex

Axis of Symmetry

y -

intercept

a > 0 ⇓ upward

(1, - 2.25)

x = 1

(0, - 2)

Area of the Square	Area of the Triangle
$f (x) = (0.5 x)^{2}$	$g (x) = \frac{1}{2} \cdot 2 \cdot (0.5 x + 2)$
$f (x) = 0.25 x^{2}$	$g (x) = 0.5 x + 2$

Area of the Square

Area of the Triangle

f (x) = (0.5 x)^{2}

g (x) = \frac{1}{2} \cdot 2 \cdot (0.5 x + 2)

f (x) = 0.25 x^{2}

g (x) = 0.5 x + 2

$x$	$f (x) = 0.25 x^{2}$	$g (x) = 0.5 x + 2$
$- 3$	$0.25 (- 3)^{2} = 2.25$	$0.5 (- 3) + 2 = 0.5$
$0$	$0.25 (0)^{2} = 0$	$0.5 (0) + 2 = 2$
$3$	$0.25 (3)^{2} = 2.25$	$0.5 (3) + 2 = 3.5$

x

f (x) = 0.25 x^{2}

g (x) = 0.5 x + 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

$y = - 0.7 x^{2} + 0.8 x + 8$
Direction	Vertex	Axis of Symmetry	$y -$ intercept
$a < 0 ⇓ downward$	$(\frac{4}{7}, \frac{2 8 8}{3 5})$	$x = \frac{4}{7}$	$(0, 8)$

y = - 0.7 x^{2} + 0.8 x + 8

Direction

Vertex

Axis of Symmetry

y -

intercept

a < 0 ⇓ downward

(\frac{4}{7}, \frac{2 8 8}{3 5})

x = \frac{4}{7}

(0, 8)

The diagram shows a blueprint of a $14$ ft by $11$ ft rectangular wall with a square window in the center. If $x$ is the side length of the square window, the function $y = 154 - x^{2}$ gives the area, in square feet, of the wall without the window.

Write a reasonable domain for the function. Write the answer as a compound inequality.

Write the range of the function. Write the answer as a compound inequality.

What is the side length of the window if the area of the wall is

90 ?

Consider the given function that gives the area in square feet of the wall without the window. y=154-x^2 In this function, x represents the side length of the square window. We will use this information to find a reasonable domain for the function. Note that it does not make sense for the side length of our window to be negative. This means that x takes only positive values. y=154-x^2, where x>0 We already know the lower bound for the domain of the function. The upper bound can be determined by using the given information about the wall. Note that the width of the wall is 11 feet, which means that x cannot be greater than 11. Domain [-0.7em] x>0 and x<11 ⇓ 0

The range of the function will be based on the domain of the function. Domain: 0points on the parabola.

x	y=135-x^2	y
0	y=154-( 0)^2	154
3	y=154-( 3)^2	145
6	y=154-( 6)^2	118
9	y=154-( 9)^2	73
11	y=154-( 11)^2	33

We can now plot these points and then connect them to finish our parabola.

By observing the graph, the y-values continue to decrease as the x-values increase, which means that the function is decreasing. In this case, the range can be determined by noting which y-values correspond to the endpoints of its domain.

x	y
0	154
11	33

This tells us that the smallest y is 33, while the largest is 154. We can now write the range of the function. Range: 33 < y < 154

We want to estimate x when y=90. To do so, we can use the graph obtained in Part B to identify the x-coordinate when y=90.

We can see from the graph that the function has a y-value of 90 when the x-value is 8. Since x represents the side length of the window, the side length of the window is 8 feet.

Complete the following statement. Suppose $a \neq = 0 .$

The graph of $y = a x^{2} + c$ intersects the $x -$ axis twice when $\underline{m m m m m m m} .$

A . B . C . D . a and c have the same sign a and c have different signs c is positive a = 1 and c = 0

Let's start by reviewing the effects of the parameters a and c in a quadratic function of the form y=ax^2+c.

If a>0, the parabola opens upwards.
If a<0, the parabola opens downwards.
If c>0, the graph is translated upwards by c units.
If c<0, the graph is translated downwards by |c| units.

Note that the parameter a can also shrink or stretch the graph of the function vertically. However, the graph is still a parabola. Remember that the vertex of the parabola y=ax^2 is at the origin, intersecting the x-axis at just one point. Let's consider the case a>0.

If we translate the same graph downwards then it intersects the x-axis twice. For this to happen, the parameters should be a>0 and c<0.

However, if the parabola opens downwards we have to translate it upwards instead. This happens when a<0 and c>0.

With what we discussed so far, we now have enough information to complete the sentence.

The graph of y=ax^2+c intersects the x-axis in two places when a andc have different signs.

Therefore, the answer is option B.

The following are known about a certain quadratic equation.

The equation has two solutions, one of which is $x = - 16 .$
The axis of symmetry of the related quadratic function is $x = - 6$
The minimum $y -$ value of the related function is $- 16 .$

Find the other solution.

It is a given that x=- 16 is a solution to the quadratic equation. Quadratic Equation: & ax^2+bx+c = 0 Solution: & x = - 16 In other words, (- 16,0) is an x-intercept of the related quadratic function. Related Quadratic Function: & y = ax^2+bx+c x-intercept: & (- 16,0) It is a given that the line x=- 6 is the axis of symmetry of the related function. Since the axis of symmetry for a quadratic function passes through the vertex and its minimum happens at its vertex, the vertex of the function is (- 6,- 16). Related Quadratic Function: & y = ax^2+bx+c x-intercept: & (- 16,0) Vertex: & (- 6, - 16) We can now trace the axis of symmetry to find the other solution. The graph of the related function opens upwards because its vertex is a minimum.

The graph of the function should be symmetric with respect to the line x=-6. Since the known x-intercept is 10 units to the left from the axis of symmetry, the other one should be 10 units to the right of it. Hence, the other root should be x= - 6+10 = 4.

Therefore, the other solution of the quadratic equation is 4.

How many different parabolas have

- 3

and

3

x -

intercepts?

Let's start by recalling the intercept form of a quadratic function. y(x) = a(x-p)(x-q) In this form, a is a nonzero real number and p and q are the x-intercepts of the function. To write a quadratic function with the x-intercepts x=3 and x=- 3, we first need to substitute p=3 and p=- 3 in the intercept form so that the output is equal to zero. p = 3 , q = - 3 ⇒ y(x) = a(x -3)(x +3) Now, we can choose a to be any nonzero real number. Let's look at some examples. y_1 = & 1 (x-3)(x+3) y_2 = & 0.5(x-3)(x+3) y_3 = & - 1 (x-3)(x+3) y_4 = & - 0.5(x-3)(x+3) Since there are infinitely many choices for a, there are infinitely many different parabolas that intersect the x-axis at x=- 3 and x=3. Some examples can be seen in the following graph.

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