Question

How do you find the vertex and intercepts for `y=8(x-10)^2-16`?

Answer 1

The vertex form of a quadratic function is given by
f (x) = a(x - h)2 + k, where (h, k) is the vertex of the parabola.

So here, it's already in vertex form. a = 8, h = 10, k = -16, and the vertex coordinates are 10, -16

Intercepts can be easily calculated. The y intercept is found when x = 0 is substituted into the original equation:

`y = 8(-10)^2 - 16 = 800 - 16 = 784`

x intercepts are found by setting y = 0 and solving for x.

`0 = 8(x - 10)^2 - 16`

...easiest way to solve this is to convert it to a standard format quadratic equation.

Multiply out:

`8(x^2 - 20x + 100) - 16 = 0`

`8x^2 - 160x + 784 = 0`

...this is standard form, `ax^2 + bx + c = 0`, with
a = 8, b = -160, c = 784

giving roots 11.41 and 8.59 (rounded)

...always good to have a graph as function as a sanity check:

graph{8(x - 10)^2 - 16 [-10.51, 21.52, -19.47, -3.45]}

GOOD LUCK!

Answer 2

Vertex is `(10,-16)`, `x`-intercept are at `(10-sqrt2,0)` and `(10+sqrt2,0)` and `y`-intercept is at `(0,784)`.

Explanation:

To find the vertex, we should have equation into vertex form i.e. `y=a(x-h)^2+k`, in which case vertex is `(h,k)`.

As `y=8(x-10)^2-16` is alreaddy in this form, with `a=8`, `h=10` and `k=-16`

Hence, vertex is `(10,-16)`

For `x`-intercepts, we should put `y=0` and for `y`-intercepts, we should put `x=0`

Putting `x=0`, `y=8(0-10)^2-16=784` and `y`-intercept is at `(0,784)`.

and when `y=0`, we have `8(x-10)^2-16=0` i.e. `(x-10)^2=2` i.e. `x=10+-sqrt2` and `x`-intercept are at `(10-sqrt2,0)` and `(10+sqrt2,0)`.

`color(red)("Graph not drawn to scale")`
graph{8(x-10)^2-16 [-5, 20, -100, 860]}