Answer:


C) See attachment.
Step-by-step explanation:
Given function:

<h3><u>Part A</u></h3>
To factor a <u>quadratic</u> in the form
<em> , </em>find two numbers that multiply to
and sum to
:


Therefore, the two numbers are -5 and 4.
Rewrite
as the sum of these two numbers:

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

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

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

Therefore:


So the x-intercepts are:

<h3><u>Part B</u></h3>
The x-value of the vertex is:

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

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

Therefore, the vertex of the function is:

<h3><u>Part C</u></h3>
Plot the x-intercepts found in Part A.
Plot the vertex found in Part B.
As the <u>leading coefficient</u> 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 <u>x-value</u> of the <u>vertex</u>. Draw a line at x = ¹/₄ and use this to ensure the drawing of the parabola is <u>symmetrical</u>.
Draw a upwards opening parabola that has a minimum point at the vertex and that passes through the x-intercepts (see attachment).