Part A
[tex]4x^2 -7x-15=0\\\\(4x+5)(x-3)=0\\\\x=-\frac{5}{4}, 3[/tex]
So, the x-intercepts are [tex]\boxed{\left(-\frac{5}{4}, 0 \right), (3,0)}[/tex]
Part B
The vertex will be a minimum because the coefficient of [tex]x^2[/tex] is positive.
The x-coordinate of the vertex is [tex]x=-\frac{-7}{2(4)}=\frac{7}{8}[/tex]
Substituting this back into the function, we get [tex]f\left(\frac{7}{8} \right)=4\left(\frac{7}{8} \right)^2 -7\left(\frac{7}{8} \right)^2 -15=-\frac{289}{16}[/tex]
So, the coordinates of the vertex are [tex]\boxed{\left(\frac{7}{8}, -\frac{289}{16} \right)}[/tex]
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.
