Functions and Graphs - Algebra 2

The first step of questions like this is to get the quadratic in its standard form. So we move the over to the left side of the inequality:

This quadratic can easily be factored as. So now we can write this in the form

and look at each of the factors individually. Recall that a negative number times a negative is a positive number. Therefore the boundaries of our solution interval is going to be when both of these factors are negative. is negative whenever , and is negative whenever . Since , one of our boundaries will be . Remember that this will be an open interval since it is less than, not less than or equal to.

Our other boundary will be the other point when the product of the factors becomes positive. Remember that is positive when , so our other boundary is . So the solution interval we arrive at is

When asked to solve for x we need to isolate x on one side of the equation.

To do this our first step is to subtract 7 from both sides.

From here, we divide by 4 to solve for x.

When asked to solve for y we need to isolate the variable on one side and the constants on the other side.

To do this we first add 9 to both sides.

From here, we divide by -12 to solve for y.

The region contains only values which are greater than or equal to those on the line , so its values are .

Similarly, the region contains only values which are less than or equal to those on the line , so its values are .

The region contains only values which are greater than or equal to those on the line , so its values are .

Also, the region contains only values which are less than or equal to those on the line , so its values are .

To begin, we analyze the equation given: the base equation, is shifted left one unit and vertically stretched by a factor of 2. The graph of the equation is:

To solve the inequality, we need to take a test point and plug it in to see if it matches the inequality. The only points that cannot be used are those directly on our parabola, so let's use the origin . If plugging this point in makes the inequality true, then we shade the area containing that point (in this case, outside the parabola); if it makes the inequality untrue, then the opposite side is shaded (in this case, the inside of the parabola). Plugging the numbers in shows:

Simplified as:

Which is not true, so the area inside of the parabola should be shaded, resulting in the following graph:

The parabola of an equation of the form is vertical, and faces upward or downward depending entirely on the sign of , the coefficient of . This coefficient, , is negative; the parabola is concave downward.

The graph of the quadratic function is a parabola with its vertex at the point with coordinates

.

Set ; the -coordinate is

Evaluate by substitution:

The vertex has 0 as its -coordinate; it is therefore on an axis.

The graph of an equation of the form

is a horizontal parabola. Whether it is concave to the left or to the right depends on the sign of . Since , a negative number, the parabola is concave to the left.

We have the following answer choices.

The first equation is a cubic function, which produces a function similar to the graph. The second equation is quadratic and thus, a parabola. The graph does not look like a prabola, so the 2nd equation will be incorrect. The third equation describes a line, but the graph is not linear; the third equation is incorrect. The fourth equation is incorrect because it is an exponential, and the graph is not an exponential. So that leaves the first equation as the best possible choice.

Starting with

moves the parabola by units to the right.

Similarly moves the parabola by units to the left.

Hence the correct answer is option .

The highest exponent of the variable term is two (). This tells that this function is quadratic, meaning that it is a parabola.

The graph below will be the answer, as it shows a parabolic curve.

To find where the graph crosses the horizontal axis, we need to set the function equal to 0, since the value at any point along the axis is always zero.

To find the possible rational zeroes of a polynomial, use the rational zeroes theorem:

Our constant is 10, and our leading coefficient is 1. So here are our possible roots:

Let's try all of them and see if they work! We're going to substitute each value in for using synthetic substitution. We'll try -1 first.

Looks like that worked! We got 0 as our final answer after synthetic substitution. What's left in the bottom row helps us factor down a little farther:

We keep doing this process until is completely factored:

Thus, crosses the axis at .

The graph of a quadratic function has an -intercept at any point at which , so, first, set the quadratic expression equal to 0:

The number of -intercepts of the graph is equal to the number of real zeroes of the above equation, which can be determined by evaluating the discriminant of the equation, . Set , and evaluate:

The discriminant is equal to zero, so the quadratic equation has one real zero, and the graph of has exactly one -intercept.

crosses the axis when equals 0. So, substitute in 0 for :

The zeros of the parabola are at and , so when placed into the formula

,

each of their signs is reversed to end up with the correct sign in the answer. The coefficient can be found by plugging in any easily-identifiable, non-zero point to the above formula. For example, we can plug in which gives

When determining the maximum number of turns a polynomial function might have, one must remember:

Max Number of Turns for Polynomial Function = degree - 1

First, we must find the degree, in order to determine the degree we must put the polynomial in standard form, which means organize the exponents in decreasing order:

Now that f(x) is in standard form, the degree is the largest exponent, which is 8.

We now plug this into the above:

Max Number of Turns for Polynomial Function = degree - 1

Max Number of Turns for Polynomial Function = 8 - 1

which is 7.

The correct answer is 7.

In order to determine the end behavior of a polynomial function, it must first be rewritten in standard form. Standard form means that the function begins with the variable with the largest exponent and then ends with the constant or variable with the smallest exponent.

For f(x) in this case, it would be rewritten in this way:

When this is done, we can see that the function is an Even (degree, 4) Negative (leading coefficent, -3) which means that both sides of the graph go down infinitely.

In order to answer questions of this nature, one must remember the four ways that all polynomial graphs can look:

Even Positive:

Even Negative:

Odd Positive:

Odd Negative:

The way to figure out this problem is by understanding behavior of polynomials.

The sign that occurs before the is positive and therefore it is understood that the function will open upwards. the "8" on the function is an even number which means that the function is going to be u-shaped. The only answer choice that fits both these criteria is: