Question

ready
a golf - pro practices his swing by driving golf balls off the edge of a cliff into a lake. the height of the ball above the lake (measured in meters) as a function of time (measured in seconds and represented by the variable t) from the instant of impact with the golf club is
58.8 + 19.6t-4.9t².
the expressions below are equivalent:
-4.9t² + 19.6t + 58.8 standard form
-4.9(t - 6)(t + 2) factored form
-4.9(t - 2)²+78.4 vertex form

1. which expression is the most useful for finding how many seconds it takes for the ball to hit the water? why?
2. which expression is the most useful for finding the maximum height of the ball? justify your answer.
3. if you wanted to know the height of the ball at exactly 3.5 seconds, which expression would help the most to find the answer? why?
4. if you wanted to know the height of the cliff above the lake, which expression would you use? why?

set
topic: finding multiple representations of a quadratic
one form of a quadratic function is given. fill - in the missing forms.
5 a. standard form b. vertex form c. factored form y=(x + 5)(x - 3)
d. table (include the vertex and at least 2 points on each side of the vertex.) e. graph
show the first differences and the second differences.

Explanation:

1.

Step1: Analyze ball - hitting - water situation

When the ball hits the water, height \(h = 0\). The factored form \(-4.9(t - 6)(t + 2)=0\) can be easily solved for \(t\) using the zero - product property (\(a\times b = 0\) implies \(a = 0\) or \(b = 0\)). So \(t-6=0\) gives \(t = 6\) and \(t + 2=0\) gives \(t=-2\). Since time \(t\geq0\), we take \(t = 6\).

2.

Step1: Analyze maximum - height situation

The vertex form \(-4.9(t - 2)^2+78.4\) is in the form \(y=a(x - h)^2 + k\), where the vertex of the parabola is \((h,k)\). For a quadratic function \(y = ax^2+bx + c\) (\(a<0\) here, \(a=-4.9\)), the vertex represents the maximum point. So the vertex \((2,78.4)\) gives the time \(t = 2\) seconds when the maximum height of \(78.4\) meters is reached.

3.

Step1: Analyze finding height at a specific time

The standard form \(-4.9t^2+19.6t + 58.8\) is most useful. We can directly substitute \(t = 3.5\) into the expression \(-4.9(3.5)^2+19.6(3.5)+58.8\) to find the height of the ball at \(t = 3.5\) seconds.

4.

Step1: Analyze finding cliff height

The standard form \(-4.9t^2+19.6t + 58.8\) is used. At the moment of impact (\(t = 0\)), substituting \(t = 0\) into \(-4.9t^2+19.6t + 58.8\) gives \(h=-4.9(0)^2+19.6(0)+58.8 = 58.8\) meters, which is the height of the cliff above the lake.

5.

Step1: Find standard form from factored form

\[

$$\begin{align*} y&=(x + 5)(x - 3)\\ &=x^2-3x+5x-15\\ &=x^2 + 2x-15 \end{align*}$$

\]

Step2: Find vertex form from standard form

For \(y=x^2 + 2x-15\), complete the square. \(y=(x^2+2x+1)-1 - 15=(x + 1)^2-16\)

Step3: Create a table

For \(y=(x + 1)^2-16\), the vertex is \((-1,-16)\).
When \(x=-3\), \(y=(-3 + 1)^2-16=4 - 16=-12\)
When \(x=-2\), \(y=(-2 + 1)^2-16=1 - 16=-15\)
When \(x=0\), \(y=(0 + 1)^2-16=1 - 16=-15\)
When \(x=1\), \(y=(1 + 1)^2-16=4 - 16=-12\)

\(x\)	\(y\)
\(-2\)	\(-15\)
\(-1\)	\(-16\)
\(0\)	\(-15\)
\(1\)	\(-12\)

Step4: First and second differences

First differences:
\(\Delta y_1=-15-(-12)=-3\), \(\Delta y_2=-16-(-15)=-1\), \(\Delta y_3=-15-(-16)=1\), \(\Delta y_4=-12-(-15)=3\)
Second differences:
\(\Delta(\Delta y)_1=-1-(-3)=2\), \(\Delta(\Delta y)_2=1-(-1)=2\), \(\Delta(\Delta y)_3=3 - 1=2\)

Answer:

1. The factored form \(-4.9(t - 6)(t + 2)\) is most useful for finding how many seconds it takes for the ball to hit the water because we can use the zero - product property to solve for \(t\) when the height \(h = 0\).
2. The vertex form \(-4.9(t - 2)^2+78.4\) is most useful for finding the maximum height of the ball since the vertex \((h,k)\) of the parabola \(y=a(x - h)^2 + k\) gives the time \(t = h\) and the maximum height \(y = k\).
3. The standard form \(-4.9t^2+19.6t + 58.8\) is most useful for finding the height of the ball at \(t = 3.5\) seconds as we can directly substitute \(t\) into the expression.
4. The standard form \(-4.9t^2+19.6t + 58.8\) is most useful for finding the height of the cliff above the lake as we substitute \(t = 0\) into the expression.
5. a. Standard form: \(y=x^2 + 2x-15\)

b. Vertex form: \(y=(x + 1)^2-16\)
c. Factored form: \(y=(x + 5)(x - 3)\)
d. Table:

\(x\)	\(y\)
\(-2\)	\(-15\)
\(-1\)	\(-16\)
\(0\)	\(-15\)
\(1\)	\(-12\)

First differences: \(-3,-1,1,3\); Second differences: \(2,2,2\)
e. Graph: Plot the points from the table and draw a parabola with vertex \((-1,-16)\) opening upwards.