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 \(\displaystyle (6,0)\).