# SAT Math : Graphing

## Example Questions

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### Example Question #1 : How To Graph A Quadratic Function

How many times does the equation below cross the x-axis?

Explanation:

You can solve this problem two ways.

3. You can solve for where the graph crosses the x-axis by setting the equation equal to zero, factoring, and solving.

2. You can quickly sketch the graph by choosing some x values and solving for y.

We see that the graph passes the x-axis twice.

### 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

16

Explanation:

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.

### Example Question #1 : How To Graph A Quadratic Function

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 and II only

I only

I and III only

II and III 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.