Question

(08.07 HC)

An expression is shown below:
f(x) = 2x²-3x - 5
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)
(10 points)

Answer 1

Below in bold.

Explanation:

Part A

At the x-intercepts f(x) = 0, therefore:

2x^2 - 3x - 5 = 0

(2x - 5)(x + 1) = 0

x = 5/2, -1)

So the x intercepts are (-1, 0) and (5/2, 0).

Part B

The Coefficient of x^2 is positive ( it is 2) So the graph opens upwards and the vertex will be a minimum.

Convert f(x) to vertex form, by completing the square:

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

= 2(x^2 - 3/2x) - 5

= 2[(x - 3/4)^2 - 9/16] - 5

= 2(x - 3/4)^2 -18/16 - 80/16

= 2(x - 3/4)^2 - 49/8

So the coordinates of the vertex are

(3/4, -49/8) or

(0.75, -6.125) in decimal form.

Part C

To graph f(x) you would first mark the points on the x axis which we found in Part 1 and the vertex found in Part 2. This vertex will be the bottom of the 'U'.

The graph is a parabola shaped roughly like a U, and will be symmetrical about the line x = 0.75 (which passes through the vertex).

You would also plot 2 more points above the x axis so as to get an accurate graph. 1 would be to the left of the line of symmetry and 1 to its right.

Suggest x = - 2 and calculate f(-2) = 2(-2)^2 - 3(-2) - 5 = 11.

- that is the point (-2,11) and the other would be x = 4, f(x) = 15. (4, 15)

Once you have plotted these points draw a smooth u shaped curve through them.