161 Precalculus 1Review 2
1. Consider the quadratic function.
a. Use the completion of squares to write the function in the standard form.
Solution. First we will factor out the coefficient by.
.
Next we will complete the square inside the parentheses using the formula
.
In our case,.
Therefore, and
Finally, and this is the standard form of our quadratic function.
b. Find the coordinates of the vertex, the x-intercepts (if any), and the y-intercept.
Solution. The coordinates of the vertex can be found immediately from the standard form of the function. Namely,.
The x-intercepts we can find by solving the quadratic equation. We can solve it by factoring, or using the quadratic formula, or the standard form we have found. If we make the last choice we write
Therefore the x-intercepts are.
Finally, the y-intercept equals to.
c. Graph the function.
Solution. To graph the function we can plot the vertex, the x-intercepts , and the y-intercept . To plot a good graph you are advised to plot a few more points on the graph. For example we can compute the values of the function at the points and plot the points.
2. A projectile is fired vertically up from the top of a 120 ft high tower with initial velocity 48 ft/sec.
a. When will the projectile reach its maximum height? What is this maximum height?
Solution. If an object is fired from initial height ft with initial velocity ft/sec its height at moment tis. In our case it will be. The height h is a quadratic function of t with the coefficients.
The function h takes its greatest value at the vertex. This greatest value is. Therefore the projectile will reach the maximum height after 1.5 sec. and this maximum height is 156 ft.
b. When will the projectile hit the ground?
Solution. When the projectile hits the ground we have. So, to answer the question we have to solve the equation. After dividing both parts by -8 we get. Let us solve it using the quadratic formula
Only the positive solution makes sense whence
3. Compute and write as a complex number in the standard form.
Solution. Let us first perform multiplications in the numerator and the denominator of our fraction. . Recall that whence the last fraction is equal to. Division is multiplication by the reciprocal and the reciprocal to the complex number is. In particular the reciprocal to is . Now we have to perform the multiplication. This is our answer.
4. Find all the solutionsof the polynomial equationand factor the polynomial completely.
Solution. We will use the rational root test. The numerators of possible rational roots can be only factors of 6:. The denominators are positive factors of 2:1, 2. Therefore the complete list of possible rational roots is. Let us check these numbers using synthetic division. We will start with 1/2.
1/2 / 2 / -3 / -6 / 13 / -61 / -1 / -7/2 / 19/4
2 / -2 / -7 / 19/2 / -5/4
We see that 1/2 is not a solution of our equation. Next we try -1/2.
-1/2 / 2 / -3 / -6 / 13 / -6-1 / 2 / 2 / -15/2
2 / -4 / -4 / 15 / -27/2
-1/2 is not a solution either. Next we try 1.
1 / 2 / -3 / -6 / 13 / -62 / -1 / -7 / -6
2 / -1 / -7 / 6 / 0
1 is a solution from the result of synthetic division we have the following factorization.
.
Now we have to solve the equation. The list of possible rational roots hasshrunk to. We will try 1 once again.
1 / 2 / -1 / -7 / 62 / 1 / -6
2 / 1 / -6 / 0
So 1 is the solution of our new equation and also a multiple solution of our original equation of multiplicity at least 2. Also we get the following factorization.
.
It remains to solve the quadratic equation. The quadratic formula gives us
. Therefore we have two more solutions. Notice that both of them are, of course, on our list of possible rational roots. So all the solutions of our equation are
And the complete factorization is
.
5. Solve the inequality.
Solution. The solution of the previous problem tells us that the left part of the inequality is 0 at points -2, 1, 3/2. The sign of the polynomial can change only at these points. Let us look at the following table
Interval / (-∞, -2) / (-2, 1) / (1, 3/2) / (3/2, ∞)Sign / + / - / - / +
The signs in the second row can be explained as follows. The sign to the left of -2 and to the right of 3/2 is positive because the leading term 2x4 is always non-negative. The point -2 is a simple root of our polynomial so the sign at -2 changes from + to –. The point 1 is a root of even multiplicity (two) and the sign does not change. Finally, 3/2 is again a simple root and the sign at 3/2 changes from – to +.
Another way to see it is to pick up a point inside each interval and to look at the sign of the polynomial at that point. For example, the value at 0 is -6 and therefore the sign of the polynomial on the interval (-2, 1) is negative.
Now we see from the table that the solution of the inequality is the union of two open intervals.
Notice that if the inequality were not strict, , the solution would be the closed interval [-2, 3/2].
6. Graph the polynomial.
Solution.The polynomial P is proportional to the polynomial from the previous problem. The factor 0.2 is so called scaling factor; we use it to prevent the graph from going to far down or up and to see better its features.
We see from the table in the previous problem what the x-intercepts are and what is the sign of P. So we know where the graph goes below and above the x-axis. We can compute additionally a few values of P. .
7. Solve the equation.
Solution. Let us square both parts of our equation.
.
Applying to the right part the formula we get
.
From here,
.
After dividing both parts by 2 we get
.
Let us again square both parts.
.
Or,
.
After moving all the terms to the left we have a quadratic equation for x.
.
The quadratic formula tells us that
.
So either. We must check these solutions by plugging them into original equation because when we square both parts of such an equation we can get so called false solutions. If we plug in then the left part is and the right part is, so 6 is a solution of our equation. If we plug in then the left part is but the right part is. Finally we see that the equation has only one solution.
8. Find all the solutions of the equation.
Solution. We have here an equation of the quadratic type. Indeed, the left part can be written as. Because we can write our equation as. To factor the left part completely let us recall the formulasand . Applying these formulas we see that and. Our equation can be now written as.
We can see at once two real solutions:. To find the remaining four solutions we have to solve two quadratic equations: and.
Applying the quadratic formula to the first of these equations we obtain
.
Similarly solving the second equation we get
.
Thus we have found all six solutions of our polynomial equation of degree six – two real solutions and two pairs of conjugate complex solutions.
9. Find all the solutions of the equation.
Solution. We have to consider two cases. The first one is. In this case and. Both solutions satisfy the original equation because for both of them the expression will be positive.
In the second case we have whence. This quadratic equation has only complex solutions (its discriminant equals to -12) and therefore our original equation has only two solutions.
10. Consider the rational function.
a. Find the x and y intercepts and the vertical asymptotes (if any).
Solution. To find the x intercepts we have to solve the equation. A fraction is 0 if and only if its numerator is 0, so we have to solve the equation. Because this equation has only one real solution. So the only x-intercept is 0 and the y intercept is 0 as well.
To find the vertical asymptotes we consider the points where the denominator,, is 0. These points are -1 and 1 and therefore we have two vertical asymptotes.
b. Find the intervals where and where.
Solution. We can write. The function can change sign only at the points -1, 0, and 1. The sign of the function is given by the following table
Interval / (-∞, -1) / (-1,0) / (0,1) / (1, ∞)Sign / - / + / - / +
To explain the table notice that if x < -1 then each of the factors is negative whence is negative. Each factor is simple (of multiplicity one) therefore at each of the points -1, 0, 1 the function changes sign.
c. Is the function even or odd or neither? How does it affect its graph?
Solution. The function is odd. Indeed, . Therefore the graph of is symmetric about the origin.
d. Does the graph of have a horizontal or a slant asymptote? If it does what is an equation of such an asymptote?
Solution. The graph has anoblique asymptote because the degree of the numerator is 3 and the degree of the denominator is 2 and the difference of degrees is one.
To find an equation of the oblique asymptote we will divide the numerator by the denominator using the long division.
The quotient of division is x and therefore the equation of the slant asymptote is.
e. Graph the function and its oblique asymptote in the same coordinate system.
Solution. Far left the graph is approaching the oblique asymptote. Then it will go along the vertical asymptote,, down, because the sign is negative. Between 0 and 1 the function is positive and the graph goes along the same vertical asymptote and intersects the x-axis at 0. The right part of the graph is symmetric to the left part about the origin.