# Math Help - Quadratic Graph

Draw the graph with the following equation, choosing appropriate values for x. For the graph, write down the equation of line of symmetry.

$y=2x^2-3x-4$

I couldn't factorise it, and when I completed the square I got $x=4$ or $x=-1$. Then I tried quadratic formula and got $x=\frac{3\pm\sqrt{41}}{4}$. Therefore I got the line of symmetry to be 3. However, the line of symmetry is $\frac{3}{4}$ so I must have done something wrong =\ .

2. I think that you completed the square wrongly.

$2x^2-3x-4$
$2(x^2-\frac{3}{2}x)-4$
$2(x^2-\frac{3}{2}+(-\frac{3}{4})^2-4-(-\frac{3}{4})^2$
$2(x-\frac{3}{4})^2-4-\frac{9}{16}$

hence symmetry occurs at $x=\frac{3}{4}$

Or if you use calculus

$\frac{dy}{dx}=4x-3$
Let $\frac{dy}{dx}=0$
$4x-3=0$
$4x=3$
$x=\frac{3}{4}$