A line tangent to a circle at a given point is perpendicular to the radius from the center to that point. That radius, which has endpoints $\displaystyle (0,0),(6,8)$,  has slope

$\displaystyle m = \frac{8-0}{6-0} = \frac{8}{6} = \frac{4}{3}$.

The line, being perpendicular to this radius, will have slope equal to the opposite of the reciprocal of that of the radius. This slope will be $\displaystyle -\frac{3}{4}$. Since it includes point $\displaystyle (6,8)$, we can use the point-slope form of the line to find its equation:

$\displaystyle y - y_{1} = m(x-x_{1})$

$\displaystyle y - 8= - \frac{3}{}4\left ( x-6 \right )$

$\displaystyle y - 8= - \frac{3}{4} x+ \frac{9}{2}$

$\displaystyle y - 8+ 8 = - \frac{3}{4} x+ \frac{9}{2} + 8$

$\displaystyle y = - \frac{3}{4} x+ \frac{25}{2}$

A circle is defined by an equation in the format $\displaystyle (x-h)^2+(y-k)^2=r^2$.

The center is indicated by the point $\displaystyle (h,k)$ and the radius $\displaystyle r$.

In the equation $\displaystyle (x-2)^2+(y)^2=36=6^2$, the center is $\displaystyle (2,0)$ and the radius is $\displaystyle 6$.

The line of symmetry of the parabola of the equation

$\displaystyle f(x) = ax^{2}+ bx + c$

is the vertical line

$\displaystyle x = -\frac{b}{2a}$

Substitute $\displaystyle a = 3, b = 9$:

The line of symmetry is

$\displaystyle x = -\frac{9}{2 \cdot 3} = - \frac{9}{6} = -\frac{3}{2}$

That is, the line of the equation $\displaystyle x =- \frac{3}{2}$.

The $\displaystyle x$-coordinate of the vertex of a parabola of the form 

$\displaystyle f(x) = ax^{2}+ bx + c$

is

$\displaystyle x= -\frac{b}{2a}$.

Substitute $\displaystyle a= 4, b = -16$:

$\displaystyle x= -\frac{-16}{2\cdot 4} = -\frac{-16}{8} = 2$

The $\displaystyle y$-coordinate is therefore $\displaystyle f(2)$:

$\displaystyle f(x) = 4x^{2}- 16x + 1$

$\displaystyle f(2) = 4 \cdot 2 ^{2}- 16 \cdot 2 + 1$

$\displaystyle f(2) = 4 \cdot 4- 16 \cdot 2 + 1$

$\displaystyle f(2) = 16 - 32 + 1$

$\displaystyle f(2) = -15$

We are seeking the value of $\displaystyle h$ when the graph of $\displaystyle h\left ( t\right )$ - a parabola - reaches its vertex.

To find this value, we first find the value of $\displaystyle t$. For a parabola of the equation

$\displaystyle f\left ( t\right ) = at^{2}+ bt + c$,

the $\displaystyle t$ value of the vertex is

$\displaystyle t = -\frac{b}{2a}$.

Substitute $\displaystyle a = -16, b = 60$:

$\displaystyle t = -\frac{60}{2 (-16)} = 1.875$

The height of the baseball after 1.875 seconds will be 

$\displaystyle h(1.875) = -16 \cdot 1.875^{2}+60 \cdot 1.875 + 100$

$\displaystyle h(1.875) = -56.25 +112.5 + 100 \approx 156$ feet.

The $\displaystyle x$-coordinate of the vertex of a parabola of the form 

$\displaystyle f(x) = ax^{2}+ bx + c$

is

$\displaystyle x= -\frac{b}{2a}$.

Set $\displaystyle a = 2, b = 6$:

$\displaystyle x= -\frac{6}{2\cdot 2} = -\frac{6}{4} =- \frac{3}{2}$

The $\displaystyle y$-coordinate is therefore $\displaystyle f\left (- \frac{3}{2} \right )$:

$\displaystyle f(x) = 2x^{2} + 6x + 5$

$\displaystyle f \left ( -\frac{3}{2} \right )= 2\left ( -\frac{3}{2} \right )^{2}+ 6\left ( -\frac{3}{2} \right ) + 5$

$\displaystyle = 2\left ( \frac{9}{4} \right ) + 6\left (- \frac{3}{2} \right ) + 5$

$\displaystyle = \frac{9}{2} - 9 + 5 = \frac{1}{2}$, which is the correct choice.

A quadratic function such as $\displaystyle h$ has a parabola as its graph. The high point of the parabola - the vertex - is what we are looking for.

The vertex of a function 

$\displaystyle h (t) = at^{2}+bt +c$

has as coordinates 

$\displaystyle \left ( - \frac{b}{2a} , h \left ( - \frac{b}{2a} \right ) \right )$ .

The second coordinate is the height and we are looking for this quantity. Since $\displaystyle h (t) = -16t^{2}+ 45t +150$, setting $\displaystyle a = -16, b= 45$:

$\displaystyle - \frac{b}{2a} = - \frac{45}{2(-16)} = \frac{45}{32} = 1.40625$ seconds for the ball to reach the peak.

The height of the ball at this point is $\displaystyle h ( 1.40625 )$, which can be evaluated by substitution:

$\displaystyle h (t) = -16t^{2}+ 45t +150$

$\displaystyle h (1.40625 ) = -16 (1.40625 )^{2}+ 45 (1.40625 ) +150$

$\displaystyle \approx -16 (1.9775 ) + 45 (1.40625 ) +150$

$\displaystyle \approx -31.6406 + 63.2813 + 150$

$\displaystyle \approx 181.6407$

Round this to 182 feet.

Set $\displaystyle f(x)= 0$ and solve for $\displaystyle x$:

$\displaystyle f(x) = 3x^{2}-9x -54$

$\displaystyle 3x^{2}-9x -54 = 0$

The terms have a GCF of 2, so

$\displaystyle 3\left ( x^{2}- 3x - 18 \right )= 0$

The trinomial in parentheses can be FOILed out by noting that $\displaystyle -6 +3 = -3$ and $\displaystyle -6 \times 3= -18$:

$\displaystyle 3(x-6)(x+3)=0$

And you can divide both sides by 3 to get rid of the coefficient:

$\displaystyle (x-6)(x+3)=0$

Set each of the linear binomials to 0 and solve for $\displaystyle x$:

$\displaystyle x - 6 = 0 \Rightarrow x = 6$

or

$\displaystyle x +3= 0 \Rightarrow x = -3$

The parabola has as its two intercepts the points $\displaystyle (-3,0)$ and $\displaystyle (6,0)$.

The amplitude of the graph of a sine function $\displaystyle f(x) = A \sin Bx$ is $\displaystyle A$. Here, $\displaystyle A= 2 \pi$, so this is the amplitude.

The functions in each of the choices take the form of a cosine function 

$\displaystyle f(x)= A \cos Nx$.

The graph of a cosine function in this form has amplitude $\displaystyle A$. Therefore, for this function to have amplitude 4, $\displaystyle A = 4$. Of the five choices, only 

$\displaystyle f(x) = 4 \cos \frac{\pi x}{5}$

matches this description.