An expression is shown below f(x)=2x^2-3x-5 part a: what are the x intercepts of the graph of f(x)? show your work. part b: is the vertex of the graph f(x) going to be a minimum or maximum? what are the coordinates of the vertex? justify your answers and show your work. 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 b to draw the graph

Given the expression for f(x)=2x^2-3x-5
a] The x-intercepts will be:
At x-intercept, f(x)=0, thus setting f(x)=0 we have:
2x^2-3x-5=0
solving the expression we get:
2x^2+2x-5x-5=0
2x(x+1)-5(x+1)=0
(2x-5)(x+1)=0
thus the x-intercept will be at ()
x=2/5 and x=-1

b] The vertex of the graph will be evaluated as follows:
the vertex form of the equation is given by:
f(x)=(x-k)^2+h
where:
(k,h) is the vertex
thus from the expression f(x)=2x^2-3x-5, the minimum or the maximum point will be at:
x=-b/2a
x=(-(-3)/(2*2))
x=3/4
the value of y will therefore be:
y=f(x)=2(3/4)^2-3(3/4)-5=-49/8
thus the vertex is at:

c] Now that we have the x-intercepts, we know where the graph will cross the x-axis, and the vertex gives us the minimum point and hence we can use them to sketch the graph. The graph will be as shown in the attachment.