How do you find the vertex of  y = 4x^2 – 6x – 3?

1 Answer
Jul 7, 2017

The vertex is the point at which the gradient is equal to $0$.

The vertex is the point $\left(\frac{3}{4} , - \frac{21}{4}\right)$

Explanation:

The vertex is the point at which the first derivative (the gradient) is equal to $0$.

$y = 4 {x}^{2} - 6 x - 3$
$y ' = 8 x - 6$

$8 x - 6 = 0 \to x = \frac{6}{8} = \frac{3}{4}$

Substitute this x-value back into the original equation to find the y coordinate of the vertex:

$y = 4 {x}^{2} - 6 x - 3 = 4 {\left(\frac{3}{4}\right)}^{2} - 6 \left(\frac{3}{4}\right) - 3$

$y = \frac{36}{16} - \frac{18}{4} - 3 = \frac{9}{4} - \frac{18}{4} - \frac{12}{4} = - \frac{21}{4}$

So the vertex is the point $\left(\frac{3}{4} , - \frac{21}{4}\right)$