Question

(08.07 HC)

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

`\textsf{A)} \quad x=-2, \:\:x=(5)/(2)`

`\textsf{B)} \quad \left((1)/(4),-(81)/(8)\right)=(0.25,-10.125)`

C) See attachment.

Explanation:

Given function:

`f(x)=2x^2-x-10`

Part A

To factor a quadratic in the form
`ax^2+bx+c` , find two numbers that multiply to
`ac` and sum to
`b` :

`\implies ac=2 \cdot -10=-20`

`\implies b=-1`

Therefore, the two numbers are -5 and 4.

Rewrite
`b` as the sum of these two numbers:

`\implies f(x)=2x^2-5x+4x-10`

Factor the first two terms and the last two terms separately:

`\implies f(x)=x(2x-5)+2(2x-5)`

Factor out the common term (2x - 5):

`\implies f(x)=(x+2)(2x-5)`

The x-intercepts are when the curve crosses the x-axis, so when y = 0:

`\implies (x+2)(2x-5)=0`

Therefore:

`\implies (x+2)=0 \implies x=-2`

`\implies (2x-5)=0 \implies x=(5)/(2)`

So the x-intercepts are:

`x=-2, \:\:x=(5)/(2)`

Part B

The x-value of the vertex is:

`\implies x=(-b)/(2a)`

Therefore, the x-value of the vertex of the given function is:

`\implies x=(-(-1))/(2(2))=(1)/(4)`

To find the y-value of the vertex, substitute the found value of x into the function:

`\implies f\left((1)/(4)\right)=2\left((1)/(4)\right)^2-\left((1)/(4)\right)-10=-(81)/(8)`

Therefore, the vertex of the function is:

`\left((1)/(4),-(81)/(8)\right)=(0.25,-10.125)`

Part C

Plot the x-intercepts found in Part A.

Plot the vertex found in Part B.

As the leading coefficient of the function is positive, the parabola will open upwards. This is confirmed as the vertex is a minimum point.

The axis of symmetry is the x-value of the vertex. Draw a line at x = ¹/₄ and use this to ensure the drawing of the parabola is symmetrical.

Draw a upwards opening parabola that has a minimum point at the vertex and that passes through the x-intercepts (see attachment).