x²   + 2x - 1 = 7

x²   + 2x - 8 = 0

(x + 4) (x - 2) = 0

x = -4, x = 2

The vertex form of a parabola is given by the equation:

\(\displaystyle f(x)=a(x-h)^2 +k\), where the point \(\displaystyle \dpi{100} (h,k)\) is the vertex, and \(\displaystyle \dpi{100} a\) is a constant.

We are told that the vertex must occur at \(\displaystyle \dpi{100} (3,4)\), so let's plug this information into the vertex form of the equation. \(\displaystyle \dpi{100} h\) will be 3, and \(\displaystyle \dpi{100} k\) will be 4.

\(\displaystyle f(x)=a(x-3)^2 +4\)

Let's now expand \(\displaystyle (x-3)^2\) by using the FOIL method, which requires us to multiply the first, inner, outer, and last terms together before adding them all together.

\(\displaystyle (x-3)^2 = (x-3)(x-3)=x^2-3x-3x+9=x^2-6x+9\)

We can replace \(\displaystyle (x-3)^2\) with \(\displaystyle x^2-6x+9\).

\(\displaystyle f(x)=a(x-3)^2+4=a(x^2-6x+9)+4\)

Next, distribute the \(\displaystyle \dpi{100} a\).

\(\displaystyle a(x^2-6x+9)+4 = ax^2 -6ax+9a+4\)

Notice that in all of our answer choices, the first term is \(\displaystyle -2x^2\). If we let \(\displaystyle \dpi{100} a=-2\), then we would have \(\displaystyle -2x^2\) in our equation. Let's see what happens when we substitute \(\displaystyle \dpi{100} -2\) for \(\displaystyle \dpi{100} a\).

\(\displaystyle f(x)=ax^2-6ax+9a+4=(-2)x^2-6(-2)x+9(-2)+4\)

\(\displaystyle =-2x^2+12x-18+4\)

We take a polynomial in the form \(\displaystyle \small ax^2 + bx +c = 0\)

and enter the corresponding coefficients into the quadratic equation.

\(\displaystyle \small \small \frac{-b\pm \sqrt{b^2 - 4ac}}{2a} = x\).  We normally expect to have two answers given by the sign \(\displaystyle \small \pm\).

So,

\(\displaystyle \small \small \small \small x^{2} + 3x - 3 = 0\) 

\(\displaystyle \small \small \small \small \frac{-(3)\pm \sqrt{(3)^2 - 4(1)(-3)}}{2(1)} = x\)

\(\displaystyle \small \small \small \small \frac{-3\pm \sqrt{9 + 12}}{2} = x\)

\(\displaystyle \small \small \small \small \small \frac{-3\pm \sqrt{21}}{2} = x\)

\(\displaystyle f(x) = x^{2} - 10x + 7\)

\(\displaystyle f(5) = 5^{2} - 10 \cdot 5 + 7\)

\(\displaystyle = 25 - 50 + 7\)

\(\displaystyle = -18\)

We solve for \(\displaystyle a\) in the equation

\(\displaystyle f (a) = f(5)\)

\(\displaystyle f(a) = -18\)

\(\displaystyle a^{2} - 10a+ 7 = -18\)

\(\displaystyle a^{2} - 10a+ 25= 0\)

\(\displaystyle (a-5)^{2} = 0\)

\(\displaystyle a-5 = 0\)

\(\displaystyle a=5\)

This is the only solution. Since it is established that \(\displaystyle a\) is not equal to 5, the correct response is that no such value exists.

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

\(\displaystyle f(9) = 9^{2} - 6 \cdot 9 + 5\)

\(\displaystyle = 81 - 54 + 5\)

\(\displaystyle = 32\)

Solve for \(\displaystyle a\) in the equation

\(\displaystyle f (a) = f(9)\)

\(\displaystyle f(a) = 32\)

\(\displaystyle a^{2} - 6a + 5 = 32\)

\(\displaystyle a^{2} - 6a -27 = 0\)

\(\displaystyle (a-9)(a+3) = 0\)

Either \(\displaystyle a- 9 = 0\), in which case \(\displaystyle a = 9\), which is already established to be untrue, or \(\displaystyle a + 3 = 0\), in which case \(\displaystyle a = -3\). This is the correct response.

This can be solved by first finding the first coordinate (\(\displaystyle t\) value) of the vertex of the parabola representing the function. 

The \(\displaystyle t\)-coordinate of the vertex is \(\displaystyle -\frac{b}{2a}\), where \(\displaystyle a = -16, b = 90\);

\(\displaystyle -\frac{b}{2a} = -\frac{90}{2(-16) } = 2.8\)

The ball takes 2.8 seconds to reach its peak. The height at that time is \(\displaystyle h (2.8)\), which is evaluated using substitution:

\(\displaystyle h\left ( t\right )= -16t^{2}+ 90 t + 600\)

\(\displaystyle h\left ( 2.8 \right )= -16\cdot 2.8^{2}+ 90 \cdot 2.8+ 600\)

\(\displaystyle = -16\cdot 7.84+ 90 \cdot 2.8+ 600\)

\(\displaystyle = -125.44+ 252+ 600\)

\(\displaystyle = 726.56\)

making the correct response 727 feet.

Set the height function equal to 0:

\(\displaystyle h\left ( t\right )=0\)

\(\displaystyle -16t^{2}+ 90 t + 600 = 0\)

Set \(\displaystyle a = -16, b = 90, c = 600\) in the quadratic formula

\(\displaystyle t = \frac{-b \pm \sqrt{b^{2}-4ac}}{2a}\)

\(\displaystyle t = \frac{-90 \pm \sqrt{90^{2}-4(-16)(600)}}{2 (-16)}\)

\(\displaystyle t = \frac{-90 \pm \sqrt{8,100+38,400}}{-32}\)

\(\displaystyle t = \frac{-90 \pm \sqrt{46,500}}{-32}\)

\(\displaystyle t \approx \frac{-90 \pm 215.64}{-32}\)

Evaluate separately, and select the positive value.

\(\displaystyle t \approx \frac{-90 +215.64}{-32} \approx \frac{125.64}{-32} \approx -3.9\)

which is thrown out.

\(\displaystyle t \approx \frac{-90 -215.64}{-32} \approx \frac{-305.64}{-32} \approx 9.6\)

which is positive and is the result we keep.

The correct response is 9.6 seconds.

The question is essentially asking for the positive value of time \(\displaystyle t\) when 

\(\displaystyle h(t) = 600\).

\(\displaystyle -16t^{2}+ 90 t + 600 = 600\)

\(\displaystyle -16t^{2}+ 90 t = 0\)

\(\displaystyle -2 t \left (8t - 45 \right ) = 0\)

Either \(\displaystyle t = 0\) - but this simply reflects that the initial height was 600 feet - or:

\(\displaystyle 8t-45 = 0\)

\(\displaystyle 8t=45\)

\(\displaystyle t = 45 \div 8 \approx 5.6\).

The ball returns to a height of 600 feet after 5.6 seconds.

The time at which the ball reaches its peak can be found by finding the \(\displaystyle t\)-coordinate of the vertex of the parabola representing the function. 

The \(\displaystyle t\)-coordinate of the vertex is \(\displaystyle -\frac{b}{2a}\), where \(\displaystyle a = -16, b = 90\);

\(\displaystyle -\frac{b}{2a} = -\frac{90}{2(-16) } = 2.8\)

The correct response is 2.8 seconds.