# How do you use the important points to sketch the graph of y=3x^2+5x-2?

Sep 5, 2017

The important points for sketching a quadratic are where it crosses or turns at the $x$ axis, crosses the $y$ axis and the vertex of the parabola.

It crosses the $x$ axis at the roots:

$3 {x}^{2} + 5 x - 2 = 0$

Factoring:

$\left(3 x - 1\right) \left(x + 2\right) = 0 \implies x = \frac{1}{3} \mathmr{and} x = - 2$

It crosses the $y$ axis where $x = 0$:

$3 {\left(0\right)}^{2} + 5 \left(0\right) - 2 = 0 \implies y = - 2$

Vertex is at$\left(- \frac{5}{6} , - \frac{47}{36}\right)$

To find vertex of $3 {x}^{2} + 5 x - 2$:

Place a bracket around the terms containing the variable.

(3x^2 + 5x)- 2 factor out the coefficient of ${x}^{2}$ if this is not 1
$3 \left({x}^{2} + \frac{5}{3} x\right) - 2$
Add the square of half the coefficient of $x$ inside the bracket and subtract it outside the bracket.
$3 \left({x}^{2} + \frac{5}{3} x + {\left(\frac{5}{6}\right)}^{2}\right) - {\left(\frac{5}{6}\right)}^{2} - 2$

Arrange
$\left({x}^{2} + \frac{5}{3} x + {\left(\frac{5}{6}\right)}^{2}\right)$ into the square of a binomial and simplify$- {\left(\frac{5}{6}\right)}^{2} - 2$:

$3 {\left(x + \frac{5}{6}\right)}^{2} - \frac{47}{36}$

This is now in the form $a {\left(x - h\right)}^{2} + k$

Where $h$ is the axis of symmetry and $k$ is the maximum or minimum value of the function.

So vertex is $\left(- \frac{5}{6} , - \frac{47}{36}\right)$

Points for drawing graph are:

$\left(\frac{1}{3} , 0\right) \left(- 2 , 0\right)$ roots of equation.

$\left(0 , - 2\right)$ $y$ axis intercept.

$\left(- \frac{5}{6} , - \frac{47}{36}\right)$ vertex