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

1 Answer
Guilherme N.
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(0)^2+7(0)-4#, then when #x=0#, #y=-4#

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

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

#(-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)#

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