Question

How do you find the vertex and intercepts for `f(x) = -9x^2 + 7x + 9`?

Answer 1

Vertex is `(7/18,13 13/36)`, zeroes are approximately `-.684 or 1.462`

Explanation:

If you write a quadratic `f(x)=ax^2+bx+c`, you can find the `x` coordinate of the vertex using the formula `(-b)/(2a)`. Let's plug our numbers into this to find the `x`.

`"-7"/"2(-9)" rArr "-7"/"-18" = 7/18`

Next, we plug `x` into `f(x)` to find the `y` of our vertex.

`f(7/18)=-9(7/18)^2+7(7/18)+9`
`f(7/18)=13 13/36`

So your vertex is `(7/18,13 13/36)`

For your zeroes, we can use the quadratic formula.

`x_(1,2) = (-b+-sqrt(color(blue)(b^2-4ac)))/(2a)`

Let's first solve our discriminate, the numbers marked in blue above.

`color(blue)(b^2-4ac) rArr (7)^2-4(-9)(9) = 373`

Now we can plug this into our quadratic formula, along with the other numbers.

`x_(1,2) = (-(7)+-sqrt(373))/(2(-9))`

`x_1 = (-7 + sqrt(373))/-18 ~~ - .684`

`x_2 = (-7 - sqrt(373))/-18 ~~ 1.462`