# How do you find the vertex and intercepts for y = 3x^2 + 12x + 5?

May 17, 2016

Vertex

The vertex is easily determined by completing the square.

$y = 3 \left({x}^{2} + 4 x + p\right) + 15$

$p = {\left(\frac{b}{2}\right)}^{2}$

$p = {\left(\frac{4}{2}\right)}^{2}$

$p = 4$

$y = 3 \left({x}^{2} + 4 x + 4 - 4\right) + 15$

$y = 3 \left({x}^{2} + 4 x + 4\right) - 4 \left(3\right) + 15$

$y = 3 {\left(x + 2\right)}^{2} - 12 + 15$

$y = 3 {\left(x + 2\right)}^{2} + 3$

In vertex form, $y = a {\left(x - p\right)}^{2} + q$, the vertex is located at the point $\left(p , q\right)$

Therefore, the vertex is at $\left(- 2 , 3\right)$
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Intercepts:

y intercept:

$y = 3 {\left(0\right)}^{2} + 12 \left(0\right) + 5$

#y = 5 -> (0, 5)

x intercept:

$0 = 3 {\left(x + 2\right)}^{2} + 3$

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

$- 1 = {\left(x + 2\right)}^{2}$

$\emptyset = x$

Therefore, there is no x intercept.

The graphical representation of this function proves that there is no x intercept.

graph{y = 3(x + 2)^2 + 3 [-22, 21.99, -11, 11]}

We could have also found, by intuition, that there are no x intercepts: In a quadratic function $y = a {\left(x - p\right)}^{2} + q$, the parameter a represents the breadth and the direction of opening of the parabola. If $a > 0$, then the parabola opens up. This is our case. Since the vertex was at $\left(- 2 , 3\right)$, which is located in the second quadrant (above the x axis), and the parabola opens up, the function will never cross the x axis.

I hope you have learned a lot and that my answer was of great help.