Question

How do you graph `y=-4(x+2)^2-1`?

Answer 1

Vertex `color(blue)(= (-2,-1))`

Axis of Symmetry `color(blue)(x = -2)`

Coefficient of `color(blue)((x^2) >0)`

Hence. the parabola opens down.

y-intercept: `x = (0, -17)`

Explanation:

We are given the function `color(red)(" "y = -4(x + 2)^2 -1)`

`color(red)(y = -4(x + 2)^2 -1)` is in Vertex Form

Note.A:-

Standard Form: `color(blue)(" "ax^2 + bx + c)`

Note.B:-

Vertex Form: `color(blue)(" "a(x-h)^2 + k)`

Vertex Form represents the Parabola `color(red)(" "y = ax^2`, translated horizontally `color(blue)h` units and vertically `color(blue)k` units

Note.C:-

Axis of Symmetry:`color(blue)(" " x = h)`

Note.D:-

In the Standard Form, if the coefficient of `x^2,` `color(blue)(a > 0)` then the parabola opens up.

In the Standard Form, if the coefficient of `x^2,` `color(blue)(a < 0)` then the parabola opens down.

Note.E:-

If the parabola opens up, we have a Minimum

If the parabola opens down, we have a Maximum

From now on, we will analyze our problem:

We are given the function `color(red)(" "y = -4(x + 2)^2 -1)`

Since the coefficient of the `color(red)(x^2)` term is greater than zero,

the parabola opens down.

In our problem, with reference to the Vertex Form,

`color(blue)(a = -4; h = -2; k=-1)`

Vertex `color(blue)( = (-2, -1)` and the maximum value is `color(red)(-1)`

Axis of Symmetry: `color(blue)( x = -2)`

To find the y-intercept substitute the value `color(red)(x = 0)` in the function

`color(red)(y = -4(x + 2)^2 -1)`

`color(blue)(y = -4(0+2)^2 - 1)`

`color(blue)(y = -4(0+2)^2 - 1)`

`color(blue)(y = -4(2)^2 - 1)`

`color(blue)(y = -16 - 1)`

`color(blue)(y = -17)`

Hence, the y-intercept is at: `color(blue)(x = -17)`

Please refer to the graph below: