Fast ill give 40 points to whoever answers this
an expression is shown below:
f(x) = 2x2 − x − 10
part a: what are the x-intercepts of the graph of f(x)? show your work. (2 points)
part b: is the vertex of the graph of f(x) going to be a maximum or minimum? what are the coordinates of the vertex? justify your answers and show your work. (3 points)
part c: what are the steps you would use to graph f(x)? justify that you can use the answers obtained in part a and part b to draw the graph. (5 points)

f(x) = 2x^2 - x - 10

a

2x^2 - x - 10 = 0

(x+2)(2x - 5) = 0

x = -2 or x = 5/2

x intercepts: (-2,0) and (5/2,0)

b

when the coefficient on x^2 is positive we get a cup - concave up positive.   so the vertex is a minimum.

the vertex of ax^2 + bx + c is -b/2a so here that's

x = )/(2(2)) = 1/4

y = f(1/4) = 2(1/16) - 1/4 - 10 = -81/8

vertex (1/4, -81/8)

justify my answer.   let's complete the square

f(x) = 2(x^2 - (1/2) x) - 10

f(x) = 2(x^2 - (1/2) x + 1/16) - 10 - 2/16

f(x) = 2(x - 1/4)^2 - 81/8

that clearly has a minimum at x=1/4, f(1/4)=-81/8

c

we can plot the three points we have and connect them into a parabola as indicated in the figure.