Question

NO LINKS!! Please help me with these graphs. Part 1​

Answer 1

Answer:

$\textsf{9)} \quad y=(x+6)^2-4$

$\textsf{10)} \quad y=-2(x+2)^2-6$

Step-by-step explanation:

Vertex form of a parabola:  

$y=a(x-h)^2+k$  where (h, k) is the vertex

Question 9

From inspection of the graph, the vertex is (-6, -4)

$\implies y=a(x+6)^2-4$

To find $a$, substitute the coordinates of a point on the curve into the equation.

Using point (-4, 0):

$\implies a(-4+6)^2-4=0$

$\implies a(2)^2-4=0$

$\implies 4a=4$

$\implies a=1$

Therefore, the equation of the parabola in vertex form is:

$y=(x+6)^2-4$

Question 10

From inspection of the graph, the vertex is (-2, -6)

$\implies y=a(x+2)^2-6$

To find $a$, substitute the coordinates of a point on the curve into the equation.

Using point (-1, -8):

$\implies a(-1+2)^2-6=-8$

$\implies a(1)^2-6=-8$

$\implies a-6=-8$

$\implies a=-2$

Therefore, the equation of the parabola in vertex form is:

$\implies y=-2(x+2)^2-6$

Answer 2

Problem 9

The instructions aren't stated anywhere, but I'm assuming your teacher wants you to find the equation of each parabola.

The vertex here is (h,k) = (-6,-4) which you have correctly determined.

This means

$y = a(x-h)^2 + k\\\\y = a(x-(-6))^2 +(-4)\\\\y = a(x+6)^2 - 4$

Next we plug in one of the other points on the parabola. We cannot pick the vertex again. Let's pick the point (-4,0) which is one of the x intercepts. We'll do this to solve for 'a'

$y = a(x+6)^2 - 4\\\\0 = a(-4+6)^2 - 4\\\\0 = a(2)^2 - 4\\\\0 = a(4) - 4\\\\0 = 4a-4\\\\4a-4 = 0\\\\4a = 4\\\\a = 4/4\\\\a = 1$

This means

$y = a(x+6)^2 - 4\\\\y = 1(x+6)^2 - 4\\\\y = (x+6)^2 - 4$

represents the equation of the parabola in vertex form.

Answer: $y = (x+6)^2 - 4$

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Problem 10

The vertex is (h,k) = (-2,-6)

So,

$y = a(x-h)^2 + k\\\\y = a(x-(-2))^2 + (-6)\\\\y = a(x+2)^2 - 6$

Now plug in another point on the parabola like (-1,-8) and solve for 'a'

$y = a(x+2)^2 - 6\\\\-8 = a(-1+2)^2 - 6\\\\-8 = a(1)^2 - 6\\\\-8 = a(1) - 6\\\\a-6 = -8\\\\a = -8+6\\\\a = -2$

Answer:  $y = -2(x+2)^2 - 6$

For each equation, you could optionally expand things out to get it into y = ax^2+bx+c form, but I think it's fine to leave it as vertex form.