# Question #c4080

Oct 13, 2016

$y = - \frac{25}{8}$

#### Explanation:

$y = 2 {x}^{2} - x - 3$

Firstly we note that at a vertex (max/min) the tangent will be horizontal ($y$=constant), so we just need to find the coordinates of the max/min. There are two ways to do this either with or without calculus:

Method 1 (pre-calculus):
Complete the square:
$y = 2 {x}^{2} - x - 3$
$\therefore y = 2 \left\{{x}^{2} - \frac{x}{2} - \frac{3}{2}\right\}$
$\therefore y = 2 \left\{{\left(x - \left(\frac{1}{4}\right)\right)}^{2} - {\left(\frac{1}{4}\right)}^{2} - \frac{3}{2}\right\}$
$\therefore y = 2 \left\{{\left(x - \frac{1}{4}\right)}^{2} - \frac{1}{16} - \frac{3}{2}\right\}$
$\therefore y = 2 \left\{{\left(x - \frac{1}{4}\right)}^{2} - \frac{25}{16}\right\}$
So a minimum occurs when $x = \frac{1}{4}$

Method 2 (Calculus):
$y = 2 {x}^{2} - x - 3$
differentiating wrt $x$ gives;
$\frac{\mathrm{dy}}{\mathrm{dx}} = 4 x - 1$
At a max/min $\frac{\mathrm{dy}}{\mathrm{dx}} = 0 \implies 4 x - 1 = 0$
$\therefore x = \frac{1}{4}$ (as above).
We observe that as the function is a positive quadratic it will have a U shared curve, so this corresponds to a minimum

Finding the tangent:
So back to business, we now know that the min occurs when $x = \frac{1}{4}$
$x = \frac{1}{4} \implies y = 2 \left(\frac{1}{16}\right) - \frac{1}{4} - 3$
$\therefore y = \frac{1}{8} - \frac{1}{4} - 3$
$\therefore y = - \frac{25}{8}$

So the equation of the tangent at the vertex (corresponding to $x = \frac{1}{4}$) is $y = - \frac{25}{8}$