Question

How do you find the axis of symmetry and vertex point of the function: `y=16x-4x^2`?

Answer 1

See explanation: Shortcuts given

Explanation:

The `x^2` tells you that it is a quadratic and thus the horse shoe type curve.

`-x^2` tells that it is an inverted horse shoe. This is the standard `(-x)^2`

Everything else transforms it in some way.

Multiplying `(-x^2) " or "x^2` by a number greater than one makes the curve steeper.

If the equation was of form `y = -x^2 +bx` then the maximum would be at `x = (-1) times b/2`. However, this equation is of form `y= ax^2 + bx` so we have to do it differently. We would change it to `y=a(x^2 +b/ax)`. Then we may apply the approach of `x =(-1) times 1/2 times b/a`

Lets try it:

Write `y=(-4)x^2 + 16x` as:
`y= -4(x^2 - 4x)`
so the maximum should occur at`(-1/2) times (-4) =2`

Plot of actual graph:

Note that what I have shown you is a shortcut to 'completing the square' method

To find where the curve crosses the x-axis substitute y=0 in the original equation. To find the y coordinate of the vertex point substitute the found value of x in the equation. I will let you do that!