What are the important points needed to graph y=3x^2+6x-1 ?

Aug 6, 2018

Vertex: $\left(- 1 , - 4\right)$ , axis of symmetry: $x = - 1$, x-intercepts:$x \approx - 2.155 \mathmr{and} x \approx 0.155$, y-intercept:
$y = - 1$, additional points:$\left(1 , 8\right) \mathmr{and} \left(- 3 , 8\right)$

Explanation:

This is equation of parabola , so vertex , axis of symmetry,

x intercepts , y intercept , opening of parabola , additional points

on the parabola are needed to draw graph.

$y = 3 {x}^{2} + 6 x - 1 \mathmr{and} y = 3 \left({x}^{2} + 2 x\right) - 1$ or

$y = 3 \left({x}^{2} + 2 x + 1\right) - 3 - 1 \mathmr{and} 3 {\left(x + 1\right)}^{2} - 4$

This is vertex form of equation ,y=a(x-h)^2+k ; (h,k)

being vertex , here $h = - 1 , k = - 4 , a = 3$ Since $a$ is positive,

parabola opens upward and vertex is at $\left(- 1 , - 4\right)$.

Axis of symmetry is x= h or x = -1 ;

y-intercept is found by putting $x = 0$ in the equation

$y = 3 {x}^{2} + 6 x - 1 \therefore y = - 1 \mathmr{and} \left(0 , - 1\right)$

x-intercepts are found by putting $y = 0$ in the equation

$0 = 3 {\left(x + 1\right)}^{2} - 4 \mathmr{and} 3 {\left(x + 1\right)}^{2} = 4$ or

${\left(x + 1\right)}^{2} = \frac{4}{3} \mathmr{and} \left(x + 1\right) = \pm \frac{2}{\sqrt{3}} \mathmr{and} x = - 1 \pm \frac{2}{\sqrt{3}}$

or $x \approx - 2.155 \mathmr{and} x \approx 0.155$ . Additional points:

$x = = 1 \therefore y = 3 {\left(1 + 1\right)}^{2} = 8 \mathmr{and} \left(1 , 8\right)$ and

$x = = - 3 \therefore y = 3 {\left(- 3 + 1\right)}^{2} = 8 \mathmr{and} \left(- 3 , 8\right)$

graph{3x^2+6x-1 [-10, 10, -5, 5]} [Ans]