Vertex Form of Parabolas

(x² - 3²)

(x² - 3)

(x + 3)(x - 3)

(x - 3)²

(x − 9)(x + 2)

(x + 9)(x + 2)

(x − 9)(x - 2)

(x − 9)(x - 2)

(w + 6)(w - 9)

(w + 3)(w - 18)

(w + 2)(w - 27)

(w - 6)(w + 9)

(5v + 8)(2v + 1)

(v + 8)(v - 1)

(5v + 8)(2v - 1)

(10v - 8)(v + 1)

n = ±√9

n = ±3

n = ±1

No real solution

(3,0)

(-3,0)

(9,0) & (-9,0)

(-3, 0) & (3, 0)

x=3,-2

x=-3,2

x=2,3

x=3,2

True

False, because the solution and root are the same but the x-intercept and zero are different

False, because the x-intercept and root are the same but the zero and solution are different

False, they are all different

x = -2

x = -15

x = -3

x = 5

x = 3

x = -5

x = 1

x = -16

-5, 2

-5, -2

5, -2

5, 2

ax + by = c

y = a(x - h)² + k

y = ax² + bx + c

y = mx + b

the y-coordinate of the vertex

the x-coordinate of the vertex

the x-intercept

the y-intercept

(3, 3)

(1, 3)

( - 1, 3)

(3, -1)

(-1, 2)

(1, 2)

(-1, 3)

(1, 3)

y = (x - 1)² + 4

y = (x - 1)² - 4

y = (x - 1)² + 3

y = (x - 1)² + 7

y = (x + 4)² - 14

y = (x + 4)² - 6

y = (x + 4)² - 18

y = (x + 4)² + 6

y = (x + 4)² + 6

y = (x + 2)² + 6

y = (x + 4)² - 6

y = (x + 2)² - 6

0

1

2

no way to tell

0

1

2

no way to tell

$f\left(x\right)=\left(x-1\right)^2+4$

$f\left(x\right)=\left(x+4\right)^2-1$

$f\left(x\right)=-\left(x-1\right)^2+4$

$f\left(x\right)=-\left(x+4\right)^2-1$

$h\left(x\right)=\ \left(x+1\right)^2-4$

$h\left(x\right)=-\left(x+1\right)^2-4$

$h\left(x\right)=\left(x+4\right)^2-1$

$h\left(x\right)=-\left(x+4\right)^2+1$

f(x) = (x - 2)² - 1

f(x) = (x + 2)² - 1

f(x) = -(x + 2)² - 1

f(x) = -(x - 2)² - 1

y = 2(x - 4)² + 5

y = 2(x + 4)² + 5

y = -2(x - 4)² + 5

y = -2(x + 4)² + 5

f(x) = (x -5)² + 1

f(x) = (x - 1)² - 5

f(x) = (x + 1)² - 5

f(x) = (x + 5)² +1

f(x) = (x -5)² + 1

f(x) = (x - 1)² - 5

f(x) = (x + 1)² - 5

f(x) = (x + 5)² +1