# What is the graph of f(x) = 2x^2 - 3x + 7?

Sep 27, 2014

To graph a quadratic equation we first need to factorise it into a different form.

First we check what the discriminant is equal to
Where $f \left(x\right) = a {x}^{2} + b {x}^{2} + c$
$\Delta$(Discriminant)$= {b}^{2} - 4 a c$

In this case $\Delta$=${3}^{2} - 4 \cdot 2 \cdot 7$
$\Delta = - 47$

Because it is less than zero it can't be factored normally
Therefore we must use the The Quadratic Formula or Completing the Square

Here I have completed the square

$f \left(x\right) = 2 {x}^{2} - 3 x + 7$

Remove factor from ${x}^{2}$ term

$f \left(x\right) = 2 \cdot \left({x}^{2} - \frac{3}{2} x + \frac{7}{2}\right)$

Take $x$ term, half it and then square it

$f \left(x\right) = - \frac{3}{2} \to - \frac{3}{4} \to \frac{9}{16}$

Add and then subtract this number inside the equation

$f \left(x\right) = 2 \cdot \left({x}^{2} - \frac{3}{2} x + \frac{9}{16} - \frac{9}{16} + \frac{7}{2}\right)$

Combine the first three terms in a perfect square

$f \left(x\right) = 2 \cdot \left({\left(x - \frac{3}{4}\right)}^{2} - \frac{9}{16} + \frac{7}{2}\right)$

Equate left over terms

$f \left(x\right) = 2 \cdot \left({\left(x - \frac{3}{4}\right)}^{2} + \frac{47}{16}\right)$

Multiply coefficient back in

$f \left(x\right) = 2 {\left(x - \frac{3}{4}\right)}^{2} + \frac{47}{8}$

This gives a turning point of $\left(\frac{3}{4} , \frac{47}{8}\right) = \left(0.75 , 5.875\right)$
and a $y$ intercept of $2 \cdot {\left(\frac{3}{4}\right)}^{2} + \frac{47}{8}$
$= \left(0 , 7\right)$ 