# How to graph a parabola y-2=-1/16(x-1)^2?

May 18, 2015

One thing we can do is expand this function.

But first, let's just isolate $y$ as follows: $y = - \left(\frac{1}{16}\right) {\left(x - 1\right)}^{2} + 2$

Now, let's expand your product of differences:

$y = - \frac{{x}^{2} - 2 x + 1}{16} + 2 = \frac{{x}^{2} - 2 x + 33}{16}$

$y = - {x}^{2} / 16 + \frac{x}{8} - 2.0625$

Using Bhaskara we can find its roots:

$\frac{\left(- \frac{1}{8}\right) \pm \sqrt{{\left(\frac{1}{8}\right)}^{2} - 4 \left(- \frac{1}{16}\right) \left(- \frac{33}{16}\right)}}{-} \left(\frac{2}{16}\right)$

$\frac{\left(- \frac{1}{8}\right) \pm \sqrt{\frac{1}{64} - \frac{33}{64}}}{-} \frac{2}{16}$

Ends up our $\Delta$ is negative, so there are no Real roots, which means this function does not cross the axis $x$.

Alternatively, we can find two other pieces of information:

Where the function crosses the axis $y$, when, by definition, $x = 0$:

$y = 0 + 0 - 2.0625$
$y = 2.0625$ is our intercept.

Now, we can find the coordinates for our vertex $\left({x}_{v} , {y}_{v}\right)$

${x}_{v} = - \frac{b}{2 a} = \frac{- \frac{1}{8}}{-} \left(\frac{2}{16}\right) = 1$
${y}_{v} = - \frac{\Delta}{4 a} = \frac{1}{2} / - \frac{1}{4} = - 2$

Vertex = ($1 , - 2$)

Finally, as the coeficient of your ${x}^{2}$ is negative, you know that this is a parabola where the vertex indicates the point of maximum.

graph{-(x^2)/16+(x/8)-2.065 [-10, 10, -5, 5]}