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

Oct 18, 2015

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 ${\left(- x\right)}^{2}$

Everything else transforms it in some way.

Multiplying $\left(- {x}^{2}\right) \text{ or } {x}^{2}$ by a number greater than one makes the curve steeper.

If the equation was of form $y = - {x}^{2} + b x$ then the maximum would be at $x = \left(- 1\right) \times \frac{b}{2}$. However, this equation is of form $y = a {x}^{2} + b x$ so we have to do it differently. We would change it to $y = a \left({x}^{2} + \frac{b}{a} x\right)$. Then we may apply the approach of $x = \left(- 1\right) \times \frac{1}{2} \times \frac{b}{a}$

Lets try it:

Write $y = \left(- 4\right) {x}^{2} + 16 x$ as:
$y = - 4 \left({x}^{2} - 4 x\right)$
so the maximum should occur at$\left(- \frac{1}{2}\right) \times \left(- 4\right) = 2$

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!