Answer:
[tex]\text{Part A.}\\(-\frac{1}{8},0),\\(\frac{3}{2},0)\\\\\text{Part B.}\\(\frac{11}{16},\frac{169}{16})\\\\\text{Part C.}[/tex]
Draw a parabola concave down with vertex at [tex](\frac{11}{16},\frac{169}{16})[/tex]. Since the leading coefficient of the equation is -16, the parabola should appear thinner than its parent function [tex]y=x^2[/tex]. Ensure that the parabola passes through the points [tex](\(-\frac{1}{8},0)[/tex] and [tex](\frac{3}{2},0)[/tex].
Step-by-step explanation:
Part A:
The x-intercepts of a function occur at [tex]y=0[/tex]. Therefore, let [tex]y=0[/tex] and solve for all values of [tex]x[/tex]:
[tex]0=-16x^2+22x+3[/tex]
The quadratic formula states that the real and nonreal solutions to a quadratic in standard form [tex]ax^2+bx+c[/tex] is equal to [tex]x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}[/tex].
In [tex]-16x^2+22x+3[/tex], assign:
- [tex]a\implies -16[/tex]
- [tex]b\implies 22[/tex]
- [tex]c\implies 3[/tex]
Therefore, the solutions to this quadratic are:
[tex]x=\frac{-22\pm\sqrt{22^2-4(-16)(3)}}{2(-16)},\\x=\frac{-22\pm 26}{-32},\\\begin{cases}x=\frac{-22+26}{-32}=\frac{4}{-32}=\boxed{-\frac{1}{8}},\\x=\frac{-22-26}{-32}=\frac{-48}{-32}=\boxed{\frac{3}{2}}\end{cases}[/tex]
The x-intercepts are then [tex]\boxed{(-\frac{1}{8},0)}[/tex] and [tex]\boxed{(\frac{3}{2},0)}[/tex].
Part B:
The a-term is negative and therefore the parabola is concave down. Thus, the vertex will be the maximum of the graph. The x-coordinate of the vertex of a quadratic in standard form [tex]ax^2+bx+c[/tex] is equal to [tex]x=\frac{-b}{2a}[/tex]. Using the same variables we assigned earlier, we get:
[tex]x=\frac{-22}{2(-16)}=\frac{-22}{-32}=\frac{11}{16}[/tex]
Substitute this into the equation of the parabola to get the y-value:
[tex]f(11/16)=-16(11/16)^2+22(11/16)+3,\\f(11/16)=\frac{169}{16}[/tex]
Therefore, the vertex of the parabola is located at [tex]\boxed{(\frac{11}{16},\frac{169}{16})}[/tex]
