ACT Math : How to graph a quadratic function

Example Questions

Example Question #1 : How To Graph A Quadratic Function

Best friends John and Elliot are throwing javelins. The height of John’s javelin is described as f(x) = -x2 +4x, and the height of Elliot’s javelin is described as f(x) = -2x2 +6x, where x is the horizontal distance from the origin of the thrown javelin. Whose javelin goes higher?

John’s

Elliot’s

The javelins reach the same height

Insufficient information provided

Elliot’s

Explanation:

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.

Example Question #1 : How To Graph A Quadratic Function

Where does the following equation intercept the x-axis?

only

and

and

and

and

and

Explanation:

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

Example Question #3 : How To Graph A Quadratic Function

Where does the following equation intercept the -axis?

and

and

and

and

and

and

Explanation:

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

Example Question #4 : How To Graph A Quadratic Function

Where does the following equation intercept the -axis?

and

and

and

and

and

and

Explanation:

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

Example Question #671 : Sat Mathematics

Let f(x) = ax2 + bx + c, where a, b, and c are all nonzero constants. If f(x) has a vertex located below the x-axis and a focus below the vertex, which of the following must be true?

I. a < 0

II. b < 0

III. c < 0

II and III only

I and II only

I and III only

I only

I, II, and III

I and III only

Explanation:

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.

Example Question #671 : Sat Mathematics

The graph of f(x) is shown above. If f(x) = ax2 + bx + c, where a, b, and c are real numbers, then which of the following must be true:

I. a < 0

II. c < 0

III. b2 – 4ac < 0

II only

I only

II and III only

I and III only

I and II only

I only

Explanation:

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 b2 – 4ac, 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:

$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

Notice, that in order for the values of x to be real, the value of b2 – 4ac, which is under the square-root sign, must be greater than or equal to zero. If b2 – 4ac 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 b2 – 4ac < 0, and choice III cannot be correct.

Only choice I is correct.

Example Question #1 : How To Graph A Quadratic Function

What is the vertex of the function  ?

Explanation:

The -coordinate of the vertex is , where .

To get the -coordinate, evaluate .

The vertex is .

Example Question #4 : How To Graph A Quadratic Function

Which of the following functions represents a parabola that has a vertex located at (–3,4), and that passes through the point (–1, –4)?

f(x) = 2x2 + 4x – 2

f(x) = x2 + 6x + 13

f(x) = –2x2 – 12x – 14

f(x) = 2x2 – 12x – 14

f(x) = x2 – 5

f(x) = –2x2 – 12x – 14

Explanation:

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 = 4a + 4

Subtract 4 from both sides.

–8 = 4a

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(+ 3)2 + 4

= –2(x + 3)(x + 3) + 4

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

= –2(x2 + 3x + 3x + 9) + 4

= –2(x2 + 6x + 9) + 4

= –2x2 – 12x – 18 + 4

= –2x2 – 12x – 14

The answer is f(x) = –2x2 – 12x – 14.

Example Question #8 : How To Graph A Quadratic Function

Let f(x) = x2. By how many units must f(x) be shifted downward so that the distance between its x-intercepts becomes 8?

16

8

12

2

4