Question

y= 3(x-3)^2-12E) Find two more points on The Graph. You can choose what x-values to use. Write your points as coordinates x y

Answer 1

Given:

`y=3(x-3)^2-12`

The quadractic equation above is written in vertex form:

`y=a(x-h)^2+k`

Where:

(h, k) is the coordinate of the vertex of the parabola

We have

a = 3

h = 3

k = -12

Let's find the following:

A.) Identify the coefficients, a, h, and k

Comparing the equation with the vertex form, we have:

a = 3

h = 3

k = -12

B.) Identify whether the graph opens up or opens down.

If a is greater than zero, then the graph opens up

If a is less than zero, then the graph opens downwards

Here, a = 3

Since a is greater than zero, the graph opens up.

The graph of the equation opens up

C.) Find the vertex.

The coordinates of the vertex is = (h, k)

Given:

h = 3

k = -12

Therefore, the vertex is: (3, -12)

D.) Find the axis of symmetry.

The axis of symmetry is the line that passes through the vertex and the focus.

To find the axis of symmetry we have:

x = h

where h = 3

Thus, the axis of symmetry is:

x = 3

E.) Let's find two more points.

Point 1 ==> (x, y)

Let's take x = 1

Substitute 1 for x and solve for y:

`\begin{gathered} y=3\mleft(x-3\mright)^2-12 \\ \\ y=3(1-3)^2-12 \\ \\ y=3(-2)^2-12 \\ \\ y=3(4)-12 \\ \\ y=12-12 \\ \\ y=0 \end{gathered}`

When x is 1, y is 0.

Therefore, we have the point:

(x, y) ==> (1, 0)

Point 2:

Let's take x = 2

Substitute 2 for x and solve for y:

`\begin{gathered} y=3\mleft(x-3\mright)^2-12 \\ \\ y=3(2-3)^2-12 \\ \\ y=3(-1)^2-12 \\ \\ y=3(1)-12 \\ \\ y=3-12 \\ \\ y=-9 \end{gathered}`

When x is 2, y is -9.

Therefore, we have the points:

(x, y) ==> (2, -9)

ANSWER:

A.) a = 3

h = 3

k = -12

B.) The graph opens up

C.) (3, -12)

D.) x= 3

E.) (1, 0), (2, -9)