Answer:
[tex]\textsf{A)} \quad x=-2, \:\:x=\dfrac{5}{2}[/tex]
[tex]\textsf{B)} \quad \left(\dfrac{1}{4},-\dfrac{81}{8}\right)=(0.25,-10.125)[/tex]
C) Â See attachment.
Step-by-step explanation:
Given function:
[tex]f(x)=2x^2-x-10[/tex]
Part A
To factor a quadratic in the form  [tex]ax^2+bx+c[/tex] , find two numbers that multiply to [tex]ac[/tex] and sum to [tex]b[/tex] :
[tex]\implies ac=2 \cdot -10=-20[/tex]
[tex]\implies b=-1[/tex]
Therefore, the two numbers are -5 and 4.
Rewrite [tex]b[/tex] as the sum of these two numbers:
[tex]\implies f(x)=2x^2-5x+4x-10[/tex]
Factor the first two terms and the last two terms separately:
[tex]\implies f(x)=x(2x-5)+2(2x-5)[/tex]
Factor out the common term  (2x - 5):
[tex]\implies f(x)=(x+2)(2x-5)[/tex]
The x-intercepts are when the curve crosses the x-axis, so when y = 0:
[tex]\implies (x+2)(2x-5)=0[/tex]
Therefore:
[tex]\implies (x+2)=0 \implies x=-2[/tex]
[tex]\implies (2x-5)=0 \implies x=\dfrac{5}{2}[/tex]
So the x-intercepts are:
[tex]x=-2, \:\:x=\dfrac{5}{2}[/tex]
Part B
The x-value of the vertex is:
[tex]\implies x=\dfrac{-b}{2a}[/tex]
Therefore, the x-value of the vertex of the given function is:
[tex]\implies x=\dfrac{-(-1)}{2(2)}=\dfrac{1}{4}[/tex]
To find the y-value of the vertex, substitute the found value of x into the function:
[tex]\implies f\left(\dfrac{1}{4}\right)=2\left(\dfrac{1}{4}\right)^2-\left(\dfrac{1}{4}\right)-10=-\dfrac{81}{8}[/tex]
Therefore, the vertex of the function is:
[tex]\left(\dfrac{1}{4},-\dfrac{81}{8}\right)=(0.25,-10.125)[/tex]
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. Â 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).
