Question

HELP!

Answer 1

Part A

$4x^2 -7x-15=0\\\\(4x+5)(x-3)=0\\\\x=-\frac{5}{4}, 3$

So, the x-intercepts are $\boxed{\left(-\frac{5}{4}, 0 \right), (3,0)}$

Part B

The vertex will be a minimum because the coefficient of $x^2$ is positive.

The x-coordinate of the vertex is $x=-\frac{-7}{2(4)}=\frac{7}{8}$

Substituting this back into the function, we get $f\left(\frac{7}{8} \right)=4\left(\frac{7}{8} \right)^2 -7\left(\frac{7}{8} \right)^2 -15=-\frac{289}{16}$

So, the coordinates of the vertex are $\boxed{\left(\frac{7}{8}, -\frac{289}{16} \right)}$

Part C

Plot the vertex and the x-intercepts and draw a parabola that passes through these three points.

The graph is shown in the attached image.