Note that the function is expressed in vertex form, y=a(x-h)^2+k, where a, h, and k are either positive or negative numbers.
f(x)=-1/9(x-30)^2+25
Let's compare the general formula for the vertex form to our function.
Formula:f(x)=& a(x- h )^2 +k
Function:f(x)=& -1/9(x- 30)^2+25
We can see that a= - 19, h= 30, and k=25. Recall that the vertex of a quadratic function written in vertex form is the point ( h,k). Therefore, the vertex of the given equation is ( 30,25). Let's plot the vertex on a coordinate plane.
Draw the Axis of Symmetry
The axis of symmetry of a quadratic function written in vertex form is the vertical line with equation x= h. As we have already noticed, for our function, this is h= 30. Therefore, the axis of symmetry is the line x= 30.
Determine and Plot the y-intercept
Recall that the x-coordinate of the y-intercept is 0. Therefore, to find the y-intercept, we will substitute x=0 in the function and simplify.
The y-intercept of the parabola occurs at the point (0,-75).
Reflect the y-intercept Across the Axis of Symmetry
The axis of symmetry divides the parabola into two mirror images. Therefore, points on one side of the parabola can be reflected across this line. Notice that the y-intercept is 30 units away from the axis of symmetry, so there exists another point directly across from the y-intercept that is also 30 units from the axis of symmetry.
Draw the Parabola
Now, with three points plotted, the general shape of the parabola can be seen. It appears that the parabola faces downward. Since a=- 19 in the given function rule, this should be expected. To draw the parabola, we will connect the points with a smooth curve.
Recall that f(x) represents the height, so it cannot be negative. Therefore, we will exclude the part of the graph which is below the x-axis.
Domain and Range
To describe the domain and range of f, let's examine its graph.
In this case, the domain of f is represented by the x-values between the x-intercepts of the graph. As we can see above, the x-intercepts of the graph are x=15 and x=45. With this, we can describe the domain as shown below.
Domain: 15 ≤ x ≤ 45
We have stated that the height of the football cannot be negative and based on the vertex of the graph, it can reach no more than 25 yards of height. Therefore, we can also describe the range.
Range: 0 ≤ y ≤ 25
b Given that g(x)=f(x+ 5), we can tell that the graph of g(x) is a translation left 5 units of the graph of f(x).
Now, we can remove the unnecessary parts and have the graph of g(x).
Proceeding in the same way as we did in Part A, we can determine its domain and range.
Domain:& 10 ≤ x ≤ 40
Range:& 0 ≤ y ≤ 25
c Recall that x is the horizontal distance (in yards) from the kicker's goal line. In this case, the goal line is located at the origin. Therefore, to determine on which possession the kicker punts closer to his goal line, we will compare the x-intercepts of their graphs that are closer to the origin.
As we can see above, the x-intercept of the graph of g(x) is less than the x-intercept of the graph of f(x). Therefore, the kicker punts closer to his goal line on the second possession.
