PLSS SOMEONE HELP ASAP WILL GIVE BRAINLIST :(
An expression is shown below: f(x) = 2x2 − 5x + 3
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: Describe the end behavior of the graph of f(x). Explain. (2 points)

PART D:
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)

Part A: The x intercepts are x coordinate that yield an output of zero:

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

We can use the quadratic formula to solve this equation:

x_{1,2}=\dfrac{5\pm\sqrt{25-24}}{4}

So, the two solutions are

x_1=\dfrac{5+1}{4}=\dfrac{3}{2},\quad x_2=\dfrac{5-1}{4}=1[/tex]

So, the graph of the function passes trough the points (1, 0) and (3/2, 0)

Part B: A quadratic equation  ax^2+bx+c always represents a parabola. The parabola is facing up (it's "U" shaped) if a is positive, while it faces down ("n" shaped) if a is negative. In this case a is positive, so the vertex is a minimum.

We can find the x coordinate of the vertex using the formula

x=-\dfrac{b}{2a}

And plug that value to find the y coordinate:

x=-\dfrac{-5}{2\cdot 2}=\dfrac{5}{4}

y=f\left(\dfrac{5}{4}\right)=-\dfrac{1}{8}

Part C: We already did pretty much everything we could have done to plot the graph. We know the vertex and the x-intercepts, so you know that the parabola will have to pass through those points. Remember that the vertical axis of symmetry passes through the vertex, and if you want you can sample some more point to be more accurate.

answer is:

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