Graphing - ACT Math

When graphed, each equation is a parabola in the form of a quadratic. Quadratics have the form y = ax2 + bx + c, where –b/2a = axis of symmetry. The maximum height is the vertex of each quadratic. Find the axis of symmetry, and plug that x-value into the equation to obtain the vertex.

Because the graph of f(x) = x2 is symmetric about the y-axis, when we shift it downward, the points where it intersects the x-axis will be the same distance from the origin. In other words, we could say that one intercept will be (-a,0) and the other would be (a,0). The distance between these two points has to be 8, so that means that 2a = 8, and a = 4. This means that when f(x) is shifted downward, its new roots will be at (-4,0) and (4,0).

Let g(x) be the graph after f(x) has been shifted downward. We know that g(x) must have the roots (-4,0) and (4,0). We could thus write the equation of g(x) as (x-(-4))(x-4) = (x+4)(x-4) = x2 - 16.

We can now compare f(x) and g(x), and we see that g(x) could be obtained if f(x) were shifted down by 16 units; therefore, the answer is 16.

f(x) must be a parabola, since it contains an x2 term. We are told that the vertex is below the x-axis, and that the focus is below the vertex. Because a parabola always opens toward the focus, f(x) must point downward. The general graph of the parabola must have a shape similar to this:

Since the parabola points downward, the value of a must be less than zero. Also, since the parabola points downward, it must intersect the y-axis at a point below the origin; therefore, we know that the value of the y-coordinate of the y-intercept is less than zero. To find the y-coordinate of the y-intercept of f(x), we must find the value of f(x) where x = 0. (Any graph intersects the y-axis when x = 0.) When x = 0, f(0) = a(0) + b(0) + c = c. In other words, c represents the value of the y-intercept of f(x), which we have already established must be less than zero. To summarize, a and c must both be less than zero.

The last number we must analyze is b. One way to determine whether b must be negative is to assume that b is NOT negative, and see if f(x) still has a vertex below the x-axis and a focus below the vertex. In other words, let's pretend that b = 1 (we are told b is not zero), and see what happens. Because we know that a and c are negative, let's assume that a and c are both –1.

If b = 1, and if a and c = –1, then f(x) = –x2 + x – 1.

Let's graph f(x) by trying different values of x.

If x = 0, f(x) = –1.

If x = 1, f(x) = –1.

Because parabolas are symmetric, the vertex must have an x-value located halfway between 0 and 1. Thus, the x-value of the vertex is 1/2. To find the y-value of the vertex, we evaluate f(1/2).

f(1/2) = –(1/2)2 + (1/2) – 1 = –1/4 + (1/2) – 1 = –3/4.

Thus, the vertex of f(x) would be located at (1/2, –3/4), which is below the x-axis. Also, because f(0) and f(1) are below the vertex, we know that the parabola opens downward, and the focus must be below the vertex.

To summarize, we have just provided an example in which b is greater than zero, where f(x) has a vertex below the x-axis and a focus below the vertex. In other words, it is possible for b > 0, so it is not true that b must be less than 0.

Let's go back to the original question. We know that a and c are both less than zero, so we know choices I and III must be true; however, we have just shown that b doesn't necessarily have to be less than zero. In other words, only I and III (but not II) must be true.

The answer is I and III only.

Because we are given the vertex of the parabola, the easiest way to solve this problem will involve the use of the formula of a parabola in vertex form. The vertex form of a parabola is given by the following equation:

f(x) = a(x – h)2 + k, where (h, k) is the location of the vertex, and a is a constant.

Since the parabola has its vertex as (–3, 4), its equation in vertex form must be as follows:

f(x) = a(x – (–3)2 + 4 = a(x + 3)2 + 4

In order to complete the equation for the parabola, we must find the value of a. We can use the point (–1, –4), through which the parabola passes, in order to determine the value of a. We can substitute –1 in for x and –4 in for f(x).

f(x) = a(x + 3)2 + 4

–4 = a(–1 + 3)2 + 4

–4 = a(2)2 + 4

–4 = 4_a_ + 4

Subtract 4 from both sides.

–8 = 4_a_

Divide both sides by 4.

a = –2

This means that the final vertex form of the parabola is equal to f(x) = –2(x + 3)2 + 4. However, since the answer choices are given in standard form, not vertex form, we must expand our equation for f(x) and write it in standard form.

f(x) = –2(x + 3)2 + 4

= –2(x + 3)(x + 3) + 4

We can use the FOIL method to evaluate (x + 3)(x + 3).

= –2(x_2 + 3_x + 3_x_ + 9) + 4

= –2(x_2 + 6_x + 9) + 4

= –2_x_2 – 12_x_ – 18 + 4

= –2_x_2 – 12_x_ – 14

The answer is f(x) = –2_x_2 – 12_x_ – 14.

Let's examine I, II, and III separately.

Because the parabola points downward, the value of a must be less than zero. Thus, a < 0 must be true.

Next, let's examine whether or not c < 0. The value of c is related to the y-intercept of f(x). If we let x = 0, then f(x) = f(0) = a(0) + b(0) + c = c. Thus, c is the value of the y-intercept of f(x). As we can see from the graph of f(x), the y-intercept is greater than 0. Therefore, c > 0. It is not possible for c < 0. This means choice II is incorrect.

Lastly, we need to examine b_2 – 4_ac, which is known as the discriminant of a quadratic equation. According to the quadratic formula, the roots of a quadratic equation are equal to the following:

Notice, that in order for the values of x to be real, the value of b_2 – 4_ac, which is under the square-root sign, must be greater than or equal to zero. If b_2 – 4_ac is negative, then we are forced to take the square root of a negative number, which produces an imaginary (nonreal) result. Thus, it cannot be true that b_2 – 4_ac < 0, and choice III cannot be correct.

Only choice I is correct.

The answer is I only.

To determine where an equation intercepts a given axis, input 0 for either (where it intercepts the -axis) or (where it intercepts the -axis), then solve. In this case, we want to know where the equation intercepts the -axis; so we will plug in 0 for , giving:

Now solve for .

Note that in its present form, this is a quadratic equation. In this scenario, we must find two factors of 12, that when added together, equal 7. Quickly, we see that 4 and 3 fit these conditions, giving:

Solving for , we see that there are two solutions,

or

The -coordinate of the vertex is , where .

To get the -coordinate, evaluate .

The vertex is .

The x intercept of an equation is the point at which it crosses the x-axis. To find the x intercept, plug in for and solve for .

To solve for , we can factor the equation. We must find two numbers that add to equal and multiply to equal . and fit these conditions, giving:

We can set each of these equal to to find two solutions for .

and

The x intercepts occur at these values, giving the coordinates:

and

The x intercept of an equation is the point at which it crosses the x-axis. To find the x intercept, plug in for and solve for .

This equation is not easily factored, so to solve for , we can use the quadratic formula:

With the equation in the form

,

, , and .

Plugging these values into the quadratic formula, we get:

Find the two solutions for :

The x intercepts occur at these values, giving the coordinates:

and

The sign in front of each number represents which direction to go on the x or y-axis starting at the origin . If the number is positive you go right on the x-axis or up on the y-axis. If the sign is negative then you move to the left on the x-axis or down on the y-axis.

To get to , move right on the x-axis by , then move down on the y-axis by .

Using the distance formula, calculate the distance from each of these points to the origin, (0, 0). While each answer choice has coordinates that add up to seven, (-1, 8) is the coordinate pair that produces the largest distance, namely , or approximately 8.06.

We know that right of the origin represents the positive x-direction, and above the origin represents the positive y-direction. For Quadrant I, both the x-coordinate and y-coordinate must be positive, i.e. (+,+). For quadrant II, the x-coordinate is negative, while the y-coordinate is positive, i.e. (+, -). For Quadrant III, both coordinates must be negative, i.e. (-, -). For Quadrant IV, the x-coordinate is positive, and the y-coordinate is negative, i.e. (+, -).

By definition, quadrant I has positive coordinates, quadrant II has coordinates, quadrant III has negative coordinates, and quadrant IV has coordinates, as shown by the diagram below. Only one answer choice falls within quadrant II. All others lie in other coordinates or one of the axes.

- Quadrant II

- Quadrant I

- x-axis

- y-axis

- Quadrant III

In order to find the values of a graphed ordered pair, you first need to the count across the x-axis (horizontal) and then across the y-axis (vertical). The middle of the x-axis and y-axis is called the origin and represents (0,0). Any movement to the left or below the origin is negative. Any movement to the right or above the origin is positive. Therefore, the coordinate for the graphed ordered pair would be two across the x-axis in the positive direction and three across the y-axis in the positive direction, or .

To find out which quadrant a point lives in, we need to remember the qualities of each quadrant. By qualities, I mean the signs associated with the x and y values which are dependent on the quadrant the point resides in.

All coordinate pairs that are in Quadrant I will have a positive x value and a positive y value.

All coordinate pairs that are in Quadrant II will have a negative x value and a positive y value.

All coordinate pairs that are in Quadrant III will have a negative x value and a negative y value.

All coordinate pairs that are in Quadrant IV will have a positive x value and a negative y value.

For the point , we see that both the x and y coordinates are positive therefore, the point resides in Quadrant I. Another way to check, we can start from the origin and going right 4 units and up 3 units we will be in the upper-right section of the coordinate-plane, which is also known as Quadrant I.

Coordinate points are given x-coordinate first, y-coordinate second. Quadrants are numbered counter-clockwise, with Quadrant I consisting of positive x-coordinates and positive y-coordinates, Quadrant II consisting of negative x-coordinates and positive y-coordinates, Quadrant III consisting of negative x-coordinates and negative y-coordinates, and Quadrant IV consisting of positive x-coordinates and negative y-coordinates.

(3, -2), Quadrant IV is correct because it lists the coordinates in the correct order and correctly identifies the Quadrant.

(3, -2), Quadrant II is incorrect because, while it lists the coordinates in the correct order, it misidentifies the Quadrant.

(-2, 3), Quadrant IV is incorrect because, while it correctly identifies the Quadrant, it lists the coordinates in the incorrect order.

(-2, 3), Quadrant II is incorrect because it both lists the coordinates in the incorrect order and misidentifies the Quadrant.

Coordinate points are given x-coordinate first, y-coordinate second. The correct representation plots the point one unit to the left of the origin on the x-axis and four units up from the origin on the y-axis.

The sign in front of each number represents which direction to go on the x or y-axis starting at the origin . If the number is positive you go right on the x-axis or up on the y-axis. If the sign is negative then you move to the left on the x-axis or down on the y-axis.

To get to , move right on the x-axis by then move down on the y-axis by .

Coordinate points are given x-coordinate first, y-coordinate second.

(-2, 1), (2, 4) is correct because it list the coordinates of each point in the correct order.

(1, -2), (4, 2) is incorrect because it lists the coordinates of each point in the incorrect order.

(2, 4), (-2, 1) is incorrect because, while it lists the coordinates of each point correctly, it misidentifies the points.

(4, 2), (1, -2) is incorrect because it both lists the coordinates of each point in the incorrect order and misidentifies the points.

The sign in front of each number represents which direction to go on the x or y-axis starting at the origin . If the number is positive you go right on the x-axis or up on the y-axis. If the sign is negative then you move to the left on the x-axis or down on the y-axis.

To get to , move left on the x-axis by , then move up on the y-axis by .