Here are partial solutions to your second midterm exam. It would be in your best interest to take the time to figure out where you made mistakes. As always, if you have questions, please ask.
Here is a good example: . There’s no need to multiply this out.
The maximum value of this quadratic is the y-coordinate of the vertex :
Primary Equation: , Secondary Equation: . Solve for y in the Secondary Equation and then substitute into the Primary Equation:
Notice that the resulting equation is a parabola opening down. To find the dimensions that maximize the area, we need to find the x-coordinate of the vertex (as in #2) and then plug this value into any version of the Secondary Equation:
So, the dimensions that maximize the area are .
First, factor out the common factor and then factor the remaining quadratic:
So, are the zeros.
Factoring by grouping works here. Also, factor the difference of 2 squares:
So, are the zeros.
The easiest way to do this one is to let and then factor:
So, , which implies that . Therefore, . Because we had to square both sides of an equation to solve, we need to check our solutions. Both do check.
Unfortunately, this one doesn’t factor. Shit! The Rational Root Theorem tells us that the following are possible rational roots: . Using Descartes Rule of Signs, we see that there are no positive roots. So, we don’t need to check any of the positive numbers in the list. Let’s test :
Great! So, we know is a factor of f. By doing long division, we get:
So, the x-intercepts are .
Since 1 is a zero with multiplicity 2, is a factor. Doing long division by twice, we get:
Note: You could use factoring by grouping after the first long division.
To find the x-intercepts, set equal to 0 and factor (this function is quadratic in form):
So, the x-intercepts are . Since the multiplicities are all 1, the graph crosses the x-axis at each of these points. To find the y-intercept, plug in . The y-intercept is . Using these points and the fact that this is an even degree polynomial (with both “ends” pointing up), sketch the graph. This graph has no asymptotes.
To find the x-intercepts, set the numerator equal to 0 and solve for x. The x-intercept is . To find the y-intercept, plug in . The y-intercept is . The vertical asymptote is since this is the value that makes the denominator 0. By comparing the degrees of the numerator and denominator (degrees are equal, so we get the ratio of the leading coefficients), we see that the horizontal asymptote is . Use this information to sketch the graph.
You must get common denominators (or keep track of the fact that ):
The x-intercepts of the corresponding function are , and the vertical asymptote is . These 3 points divide the x-axis into 4 intervals. By picking test points, we can determine where the function is above and below the x-axis. The solution is . Note: 0 is not included because it is a vertical asymptote.