Which of the following graphs described by the function given below y=3x^2+7x+2

*Parabola that opens up

*y-intercept is 2

*x-intercepts are -1/3 and -2

*vertex is (-7/6 , -25/12)

Step-by-step explanation:

I will describe what this graph looks like.

First, the graph is quadratic because it is in the form ax² + bx + c. The function must be in the shape of a parabola. Parabolas look like a "U" shape.

Whether "a" is negative or positive represents of the parabola opens up or down. Since "3" is positive, the parabola opens up.

"c" represents the y-intercept. The y-intercept is when the graph touches the y-axis. The function must have a y-intercept of positive 2.

We can also find its x-intercepts, also called roots/zeroes, by substituting into the quadratic formula (Ignore the Â).

Using the form ax² + bx + c, we know:

a=3; b=7; c=2

Substitute into the formula.

Split the equation at the ± sign.

The graph has x-intercepts -2 and -1/3.

We can find the vertex of the graph. It is the part of the parabola that is the lowest (or highest, is it opens down).

Find the midpoint of the x-intercepts for the vertex x-coordinate:

(-1/3 - 2)/2 = (-1/3 - 6/3)/2 = (-7/3)*(1/2) = -7/6 = -1.167 = x

Substitute the vertex x-coordinate into the formula to find the "y" in vertex.

y=3x²+7x+2

y=3(-7/6)²+7(-7/6)+2

y= 147/36 + (-49/6) + 2

y= 147/36 + (-294/36) + 72/36

y= (147-294+72)/36

y = -75/36

y = -25/12 = -2.083

The vertex is (-7/6 , -25/12) OR about (-1.167, - 2.083).

black/brown

step-by-step explanation:

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