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

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Answer 1 · 2015-12-22T20:43:21

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.

The intercepts:

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

$(-7+-sqrt((7)^2-4(3)(-4)))/(2(3)) = (-7+-(sqrt(97)))/6$

Let's approximate $sqrt(97)~=9.84$

$(-7+-9.84)/6 => x_1~=0.473$ and $x_2~=-2.807$

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

The vertex has the following formulae:

$x_v=-b/(2a)=-7/6~=-1.167$

$y_v=-Delta/(4a)=-97/12~=-8.083$

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

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