Question

100 POINTS please show your work

Answer 1

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:

$f(x)=2x^2-3x-5\\\\0=2x^2-3x-5\\\\0=2x^2+2x-5x-5\\\\0=2x(x+1)-5(x+1)\\\\0=(2x-5)(x+1)$

Use the Zero Product Property to find the x-intercepts:

$0=2x-5\\5=2x\\x=\frac{5}{2}$

$0=x+1\\x=-1$

Hence, the x-intercepts for the graph of f(x) are $x=-1$ and $x=\frac{5}{2}$.

Part B

Find the x-coordinate of the vertex:

$x=-\frac{b}{2a}\\ \\x=-\frac{(-3)}{2(2)}\\\\x=\frac{3}{4}$

Find the y-coordinate of the vertex:

$f(\frac{3}{4})=2(\frac{3}{4})^2-3(\frac{3}{4})-5=-\frac{49}{8}$

Hence, the vertex is $(\frac{3}{4},-\frac{49}{8})$. 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 $f(0)=2(0)^2-3(0)-5=-5$, or $(0,-5)$ as an ordered pair. See attached graph.