CPM Homework Help : CCA Problem 11-120

Graph the parabola described by the following equation by finding its $x$ -intercepts and vertex.

$y=3x^2-10x+2$

Step 1:

First, convert the equation to graphing form.

$\frac{y}{3}=x^2-\frac{10x}{3}+\frac{2}{3}$

$\frac{y}{3}+\frac{19}{9}=x^2-\frac{10}{3}+\frac{25}{9}$

$y=3\left(x-\frac{5}{3}\right)^2-\frac{19}{3}$

Step 2:

The equation helps you find the coordinates of the vertex.

$\left(\frac{5}{3},-\frac{19}{3}\right)$

Step 3:

Substitute $0$ for $y$ to find the $x$ -intercepts.
Substitute $0$ for $x$ to find the $y$ -intercept.

$\left(\frac{5+\sqrt{19}}{3}\approx3.1,0\right)\left(\frac{5-\sqrt{19}}{3}\approx0.2,0\right)\left(0,2\right)$

Answer:

An upward parabola with a vertex at (1.7, comma negative 6.3) going through the points (0, comma 2), (0.2, comma 0) and (3.1, comma 0).