Question

Identify the vertex and the y-intercept of the graph of the function. y = –1.5(x + 2)^2

Answer 1

This is written in vertex form:-

y = a(x - b)^2 + c where (b, c) is the vertex.

Comparing the function in the question to the above:-

b = -2 and c = 0 so the vertex is at the point (-2, 0) answer

The y intercept occurs when x = 0 so here it is:-

-1.5( 0 + 2)^2 = -6

y intercept is at ( 0, -6) answer

Answer 2

The vertex = (-2, 0)

The y-intercept of the equation is -6.

Explanation:

Since, the vertex form of quadratic equation,

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

Where, (h,k) is the vertex of the equation,

Here, the given equation,

`y = -1.5(x + 2)^2`

`\implies y=-1.5(x-(-2))^2+0`

By comparing,

The vertex = (-2, 0)

Now, for y-intercept,

x = 0,

`y=-1.5(0+2)^2=-1.5(4)=-6`

Hence, y-intercept of the equation is -6.