Part A
[tex]2x^2 -x-10=0\\\\(x+2)(2x-5)=0\\\\x=-2, \frac{5}{2}[/tex]
So, the x-intercepts are (-2,0) and (5/2, 0).
Part B
The vertex will be a minimum because the leading coefficient of the quadratic is positive, meaning the graph will open up.
The vertex has an x-coordinate that is the average of the roots, which in this case is 1/4.
[tex]f(1/4)=-81/8[/tex]
So, the coordinates of the vertex are [tex]\left(\frac{1}{4}, -\frac{81}{8} \right)[/tex]
Part C
Plot the two x-intercepts and the vertex. Then, draw a parabola through these points that opens up.