Answer:
[tex]\textsf{A)} \quad \left(-\dfrac{1}{2} , 0\right) \textsf{ and } (2, 0)[/tex]
[tex]\textsf{B)} \quad \left(\dfrac{3}{4},25 \right)[/tex]
C) see explanation
Step-by-step explanation:
Given function:
[tex]f(x) =-16x^2 + 24x + 16[/tex]
Part A
To find the x-intercepts, set the function to zero, factor and solve for x:
[tex]\implies f(x)=0[/tex]
[tex]\implies -16x^2 + 24x + 16=0[/tex]
[tex]\implies -8(2x^2-3x-2)=0[/tex]
[tex]\implies 2x^2-3x-2=0[/tex]
[tex]\implies 2x^2-4x+x-2=0[/tex]
[tex]\implies 2x(x-2)+1(x-2)=0[/tex]
[tex]\implies (2x+1)(x-2)=0[/tex]
Therefore:
[tex]\implies 2x+1=0 \implies x=-\dfrac{1}{2}[/tex]
[tex]\implies x-2=0 \implies x=2[/tex]
Therefore, the x-intercepts of the graph of f(x) are
[tex]\left(-\dfrac{1}{2} , 0\right) \textsf{ and } (2, 0)[/tex]
Part B
As the leading coefficient is negative, the parabola will open downwards. Therefore, the vertex of the graph of f(x) will be a maximum.
The x-coordinate of the vertex is the midpoint of the x-intercepts:
[tex]\implies \sf midpoint=\dfrac{2+\left(-\frac{1}{2}\right)}{2}=\dfrac{3}{4}[/tex]
To find the y-coordinate of the vertex, substitute the x-value into the function:
[tex]\implies f\left(\dfrac{3}{4}\right)=-16\left(\dfrac{3}{4}\right)^2 + 24\left(\dfrac{3}{4}\right) + 16[/tex]
[tex]\implies f\left(\dfrac{3}{4}\right)=-9+18+16[/tex]
[tex]\implies f\left(\dfrac{3}{4}\right)=25[/tex]
Therefore, the coordinates of the vertex are:
[tex]\left(\dfrac{3}{4},25 \right)[/tex]
Part C
Find the y-intercept by substituting x = 0 into the function:
[tex]\implies f(0) =-16(0)^2 + 24(0) + 16=16[/tex]
Therefore, the y-intercept is (0, 16)
To graph f(x)
- Plot the vertex
- Plot the x-intercepts
- Plot the y-intercept
- Draw a parabola opening downwards with the vertex as the maximum point.
The axis of symmetry is the x-value of the vertex. Use this to help ensure the curve is symmetrical.
