Answer:
See below for answers and explanations (along with a graph)
Step-by-step explanation:
Part A
Set f(x)=0 and factor the expression by grouping:
[tex]f(x)=2x^2-3x-5\\\\0=2x^2-3x-5\\\\0=2x^2+2x-5x-5\\\\0=2x(x+1)-5(x+1)\\\\0=(2x-5)(x+1)[/tex]
Use the Zero Product Property to find the x-intercepts:
[tex]0=2x-5\\5=2x\\x=\frac{5}{2}[/tex]
[tex]0=x+1\\x=-1[/tex]
Hence, the x-intercepts for the graph of f(x) are [tex]x=-1[/tex] and [tex]x=\frac{5}{2}[/tex].
Part B
Find the x-coordinate of the vertex:
[tex]x=-\frac{b}{2a}\\ \\x=-\frac{(-3)}{2(2)}\\\\x=\frac{3}{4}[/tex]
Find the y-coordinate of the vertex:
[tex]f(\frac{3}{4})=2(\frac{3}{4})^2-3(\frac{3}{4})-5=-\frac{49}{8}[/tex]
Hence, the vertex is [tex](\frac{3}{4},-\frac{49}{8})[/tex]. We can see from the positive leading coefficient of the function that the vertex will be a minimum because the parabola will open faced-up.
Part C
You can use the x-intercepts and vertex to plot points of the graph of the function. Additionally, you can throw in the y-intercept in as well. The y-intercept, in this case, is [tex]f(0)=2(0)^2-3(0)-5=-5[/tex], or [tex](0,-5)[/tex] as an ordered pair. See attached graph.