Question

go
topic: vertex form of a quadratic equation
given the vertex form of a quadratic function, identify the vertex, intercepts, and vertical stretch of the parabola.

1. $y=(x + 2)^2-4$

a. vertex:
b. x - inter(s)
c. y - inter
d. stretch

1. $y=-3(x + 6)^2+3$

a. vertex:
b. x - inter(s)
c. y - inter:
d. stretch

1. $y=2(x - 1)^2-8$

a. vertex:
b. x - inter(s)
c. y - inter
d. stretch

1. $y = 4(x + 2)^2-64$

a. vertex:
b. x - inter(s)
c. y - inter
d. stretch

1. $y=-3(x - 2)^2+48$

a. vertex:
b. x - inter(s)
c. y - inter:
d. stretch

1. $y=(x + 6)^2-1$

a. vertex:
b. x - inter(s)
c. y - inter
d. stretch

1. did you notice that the parabolas in problems 11, 12, & 13 are the same as the ones in problems 23, 24, & 25 respectively? if you didnt, go back and compare the answers in problems 11, 12, & 13 and problems 23, 24, & 25.

prove that a. $4(x - 2)(x + 6)=4(x + 2)^2-64$
b. $-3(x + 2)(x - 6)=-3(x - 2)^2+48$
c. $(x + 5)(x + 7)=(x + 6)^2-1$

Explanation:

Step1: Recall vertex - form of quadratic function

The vertex - form of a quadratic function is $y=a(x - h)^2+k$, where the vertex is $(h,k)$.

Step2: Find vertex for $y=(x + 2)^2-4$

For $y=(x + 2)^2-4$, comparing with $y=a(x - h)^2+k$, we have $h=-2$ and $k = - 4$. So the vertex is $(-2,-4)$.

Step3: Find x - intercepts for $y=(x + 2)^2-4$

Set $y = 0$, then $(x + 2)^2-4=0$. So $(x + 2)^2=4$, which gives $x+2=\pm2$. Solving for $x$, we get $x=0$ or $x=-4$.

Step4: Find y - intercept for $y=(x + 2)^2-4$

Set $x = 0$, then $y=(0 + 2)^2-4=0$.

Step5: Find vertical stretch for $y=(x + 2)^2-4$

The coefficient of $(x + 2)^2$ is $a = 1$. So the vertical stretch is $1$.

Step6: Find vertex for $y=-3(x + 6)^2+3$

For $y=-3(x + 6)^2+3$, comparing with $y=a(x - h)^2+k$, we have $h=-6$ and $k = 3$. So the vertex is $(-6,3)$.

Step7: Find x - intercepts for $y=-3(x + 6)^2+3$

Set $y = 0$, then $-3(x + 6)^2+3=0$. So $(x + 6)^2 = 1$, which gives $x+6=\pm1$. Solving for $x$, we get $x=-5$ or $x=-7$.

Step8: Find y - intercept for $y=-3(x + 6)^2+3$

Set $x = 0$, then $y=-3(0 + 6)^2+3=-3\times36 + 3=-108 + 3=-105$.

Step9: Find vertical stretch for $y=-3(x + 6)^2+3$

The coefficient of $(x + 6)^2$ is $a=-3$. So the vertical stretch is $3$ (we consider the absolute - value for stretch factor).

Step10: Find vertex for $y=2(x - 1)^2-8$

For $y=2(x - 1)^2-8$, comparing with $y=a(x - h)^2+k$, we have $h = 1$ and $k=-8$. So the vertex is $(1,-8)$.

Step11: Find x - intercepts for $y=2(x - 1)^2-8$

Set $y = 0$, then $2(x - 1)^2-8=0$. So $(x - 1)^2=4$, which gives $x - 1=\pm2$. Solving for $x$, we get $x=3$ or $x=-1$.

Step12: Find y - intercept for $y=2(x - 1)^2-8$

Set $x = 0$, then $y=2(0 - 1)^2-8=2 - 8=-6$.

Step13: Find vertical stretch for $y=2(x - 1)^2-8$

The coefficient of $(x - 1)^2$ is $a = 2$. So the vertical stretch is $2$.

Step14: Find vertex for $y=4(x + 2)^2-64$

For $y=4(x + 2)^2-64$, comparing with $y=a(x - h)^2+k$, we have $h=-2$ and $k=-64$. So the vertex is $(-2,-64)$.

Step15: Find x - intercepts for $y=4(x + 2)^2-64$

Set $y = 0$, then $4(x + 2)^2-64=0$. So $(x + 2)^2 = 16$, which gives $x+2=\pm4$. Solving for $x$, we get $x = 2$ or $x=-6$.

Step16: Find y - intercept for $y=4(x + 2)^2-64$

Set $x = 0$, then $y=4(0 + 2)^2-64=4\times4-64=16 - 64=-48$.

Step17: Find vertical stretch for $y=4(x + 2)^2-64$

The coefficient of $(x + 2)^2$ is $a = 4$. So the vertical stretch is $4$.

Step18: Find vertex for $y=-3(x - 2)^2+48$

For $y=-3(x - 2)^2+48$, comparing with $y=a(x - h)^2+k$, we have $h = 2$ and $k = 48$. So the vertex is $(2,48)$.

Step19: Find x - intercepts for $y=-3(x - 2)^2+48$

Set $y = 0$, then $-3(x - 2)^2+48=0$. So $(x - 2)^2 = 16$, which gives $x - 2=\pm4$. Solving for $x$, we get $x=6$ or $x=-2$.

Step20: Find y - intercept for $y=-3(x - 2)^2+48$

Set $x = 0$, then $y=-3(0 - 2)^2+48=-3\times4 + 48=-12 + 48=36$.

Step21: Find vertical stretch for $y=-3(x - 2)^2+48$

The coefficient of $(x - 2)^2$ is $a=-3$. So the vertical stretch is $3$.

Step22: Find vertex for $y=(x + 6)^2-1$

For $y=(x + 6)^2-1$, comparing with $y=a(x - h)^2+k$, we have $h=-6$ and $k=-1$. So the vertex is $(-6,-1)$.

Step23: Find x - intercepts for $y=(x + 6)^2-1$

Set $y = 0$, then $(x + 6)^2-1=0$. So $(x + 6)^2=1$, which gives $x+6=\pm1$. Solving for $x$, we get $x=-5$ or $x=-7$.

Step24: Find y - intercept for $y=(x + 6)^2-1$

Set $x = 0$, then $y=(0 + 6)^2-1=36 - 1=35$.

Step25: Find vertical stretch for $y=(x + 6)^2-1$

The coefficient of $(x + 6)^2$ is $a = 1$. So the vertical stretch is $1$.

Step26: Prove $4(x - 2)(x + 6)=4(x + 2)^2-64$

First, expand the left - hand side: $4(x - 2)(x + 6)=4(x^2+6x-2x - 12)=4(x^2+4x-12)=4x^2+16x-48$.
Then, expand the right - hand side: $4(x + 2)^2-64=4(x^2 + 4x+4)-64=4x^2+16x+16-64=4x^2+16x-48$. So they are equal.

Step27: Prove $-3(x + 2)(x - 6)=-3(x - 2)^2+48$

Expand the left - hand side: $-3(x + 2)(x - 6)=-3(x^2-6x + 2x-12)=-3(x^2-4x-12)=-3x^2 + 12x+36$.
Expand the right - hand side: $-3(x - 2)^2+48=-3(x^2-4x + 4)+48=-3x^2+12x-12 + 48=-3x^2+12x+36$. So they are equal.

Step28: Prove $(x + 5)(x + 7)=(x + 6)^2-1$

Expand the left - hand side: $(x + 5)(x + 7)=x^2+7x+5x + 35=x^2+12x+35$.
Expand the right - hand side: $(x + 6)^2-1=x^2+12x+36-1=x^2+12x+35$. So they are equal.

Answer:

20.
a. $(-2,-4)$
b. $x = 0,x=-4$
c. $y = 0$
d. $1$
21.
a. $(-6,3)$
b. $x=-5,x=-7$
c. $y=-105$
d. $3$
22.
a. $(1,-8)$
b. $x = 3,x=-1$
c. $y=-6$
d. $2$
23.
a. $(-2,-64)$
b. $x = 2,x=-6$
c. $y=-48$
d. $4$
24.
a. $(2,48)$
b. $x=6,x=-2$
c. $y = 36$
d. $3$
25.
a. $(-6,-1)$
b. $x=-5,x=-7$
c. $y = 35$
d. $1$
26.
a. Proven as above.
b. Proven as above.
c. Proven as above.