How do you find the vertex and intercepts for  y = 3x^2 + 7x – 4?

Dec 22, 2015

The intercepts are given by attributing zero to one of the variables, while the vertex has its coordinates ${x}_{v}$ and ${y}_{v}$ given by formulae.

Explanation:

The intercepts:

1. When $x = 0$, $y = 3 {\left(0\right)}^{2} + 7 \left(0\right) - 4$, then when $x = 0$, $y = - 4$

Thus, the first intercept, which corsses the axle $y$ has the following coordinates: $\left(0 , - 4\right)$

1. When $y = 0$, we must resort to Bhaskara, which will go like this:

$\frac{- 7 \pm \sqrt{{\left(7\right)}^{2} - 4 \left(3\right) \left(- 4\right)}}{2 \left(3\right)} = \frac{- 7 \pm \left(\sqrt{97}\right)}{6}$

Let's approximate $\sqrt{97} \cong 9.84$

$\frac{- 7 \pm 9.84}{6} \implies {x}_{1} \cong 0.473$ and ${x}_{2} \cong - 2.807$

Thus, the intercepts that cross the axis $x$ (considering an approximation) are $\left(0.473 , 0\right)$ and $\left(- 2.807 , 0\right)$

The vertex has the following formulae:

${x}_{v} = - \frac{b}{2 a} = - \frac{7}{6} \cong - 1.167$

${y}_{v} = - \frac{\Delta}{4 a} = - \frac{97}{12} \cong - 8.083$

Vertex's coordinates: $\left(- 1.167 , - 8.083\right)$