What are the important points needed to graph y = 3x^2 + 8x - 6?

Oct 28, 2015

Its vertex is $\left(\frac{- 4}{3} , \frac{- 2}{3}\right)$
Since the co-efficient of ${x}^{2}$ is positive, the curve is open upwards.

It has a minimum at $\left(\frac{- 4}{3} , \frac{- 2}{3}\right)$
Its y- intercept is $- 6$

Explanation:

Given-

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

We have to find the vertex

$x = \frac{- b}{2 a} = \frac{- 8}{2 \times 3} = \frac{- 8}{6} = \frac{- 4}{3}$

At $x = \frac{- 4}{3}$;

$y = 3 {\left(\frac{- 4}{3}\right)}^{2} + 8 \left(\frac{- 4}{3}\right) - 6$
$y = 3 \left(\frac{16}{9}\right) - \frac{32}{3} - 6$
$y = \frac{48}{3} - \frac{32}{3} - 6 = \frac{- 2}{3}$

Its vertex is $\left(\frac{- 4}{3} , \frac{- 2}{3}\right)$

Take two points on either side of $x = \frac{- 4}{3}$

Find the y values. Plot the points. Join them with a smooth curve.

Since the co-efficient of ${x}^{2}$ is positive, the curve is open upwards.

It has a minimum at $\left(\frac{- 4}{3} , \frac{- 2}{3}\right)$

Its y- intercept is $- 6$

Since the co-efficient of ${x}^{2}$ is 3, the curve is a narrow one.

graph{3x^2+8x-6 [-25.65, 25.65, -12.83, 12.82]}