PSAT/NMSQT Prep 2022 - Eggert M.D., Strelka A. 2022

Graphs of quadratics
Quadratics
Passport to advanced math

LEARNING OBJECTIVE

After this lesson, you will be able to:

· Relate properties of a quadratic function to its graph and vice versa

To answer a question like this:

Which of the following statements is NOT true, given the equation y = (4x − 3)² + 6?

A. The vertex is (3, 6).

B. The y-intercept is (0, 15).

C. The parabola opens upward.

D. The graph does not cross the x-axis.

You need to know this:

A quadratic function is a quadratic equation set equal to y or f(x) instead of 0. Remember that the solutions (also called “roots” or “zeros”) of any polynomial function are the same as the x-intercepts. To solve a quadratic function, substitute 0 for y, or f(x), then solve algebraically. Alternatively, you can plug the equation into your graphing calculator and read the x-intercepts from the graph.

The graph of every quadratic equation (or function) is a parabola, which is a symmetric U-shaped graph that opens either upward or downward. To determine which way a parabola will open, examine the value of a in the equation. If a is positive, the parabola will open upward. If a is negative, it will open downward. Take a look at the examples above to see this graphically.

Like quadratic equations, quadratic functions will have zero, one, or two distinct real solutions, corresponding to the number of times the parabola crosses (or touches) the x- axis, as shown in the illustrations below. Graphing is a powerful way to determine the number of solutions a quadratic function has.

There are three algebraic forms that a quadratic equation can take: standard, factored, and vertex. Each is provided in the following table along with the graphical features that are revealed by writing the equation in that particular form.

You’ve already seen standard and factored forms earlier in this chapter, but vertex form might be new to you. In vertex form, a is the same as the a in standard form, and h and k are the coordinates of the vertex (h, k). If a quadratic function is not in vertex form, you can still find the x-coordinate of the vertex by plugging the appropriate values into the equation , which is also the equation for the axis of symmetry (see graph that follows). Once you determine h, plug this value into the quadratic function and solve for y to determine k, the y-coordinate of the vertex.

The equation of the axis of symmetry of a parabola is x = h, where h is the x-coordinate of the vertex.

You need to do this:

· To find the vertex of a parabola, get the function into vertex form, y = a(x − h)² + k, or use the formula

· To find the y-intercept of a quadratic function, plug in 0 for x.

· To determine whether a parabola opens upward or downward, look at the coefficient of a. If a is positive, the parabola opens upward. If negative, it opens downward.

· To determine the number of x-intercepts, set the quadratic function equal to 0 and solve or examine its graph. (Quadratic function questions show up on both the no-calculator and calculator sections of the PSAT.)

Explanation:

Be careful: the equation looks like vertex form, y = a(x − h)² + k, but it’s not quite there because the x has a coefficient of 4 inside the parentheses. You could rewrite the equation in vertex form, but this would involve squaring the quantity in parentheses and then completing the square, which would take quite a bit of time. Alternatively, you could notice that the smallest possible value for y in this function is 6, which happens when the squared term, (4x − 3)², equals zero. To check (A), find the x-value when y = 6:

So the vertex is , not (3, 6). It follows that (A) is false and is the correct answer.

For the record:

Choice (B): Substitute 0 for x and simplify to find that the y-intercept is indeed (0, 15). Eliminate.

Choice (C): The squared term is positive, so the parabola does open upward. Eliminate.

Choice (D): Because the parabola opens upward and the vertex is at y = 6, above the x-axis, the parabola cannot cross the x-axis. This statement is true as well, so it can also be eliminated, which confirms that (A) is ­correct.

Try on Your Own

Directions: Take as much time as you need on these questions. Work carefully and methodically. There will be an opportunity for timed practice at the end of the chapter.

HINT: For Q20, which form of a quadratic would you use to find its solutions?

20. The following quadratic equations are all representations of the graph shown. Which equation enables the easiest calculation of the x-intercepts of the graph?

A.

B.

C.

D.

21. Which of the following represents the function shown in the graph?

A.

B.

C.

D.

22. If the distance from −a to b in the figure shown is 10, which of the following could be the factored form of the graph’s equation?

A. y = (x − 7)(x − 3)

B.y = (x − 7)(x + 3)

C.y = (x − 8)(x − 2)

D. y = (x − 1)(x + 10)

23. If y = ax² + bx + c represents the equation of the graph shown in the figure, which of the following statements is NOT true?

A. The value of a is a negative number.

B.The value of c is a negative number.

C.The y-value is increasing for x < 3 and decreasing for x > 3.

D. The zeros of the equation are x = −2 and x = 8.

HINT: Begin Q24 by considering what the x-intercepts would represent in terms of the ball’s trajectory.

24. If a catapult is used to throw a lead ball from ground level, the path of the ball can be modeled by a quadratic equation, y = ax² + bx + c, where x is the horizontal distance that the ball travels and y is the height of the ball. If one of these catapult-launched lead balls travels 150 feet before hitting the ground and reaches a maximum height of 45 feet, which of the following equations represents its path?

A. y = −0.008x² + 1.2x

B.y = −0.008x² − 150x

C.y = 45x² + 150x

D. y = 125x² + 25x

HINT: Consider extending the graph in Q25. Don’t forget to do the unit conversion.

25. The figure above shows the partial trajectory of a cannonball shot into the air. Assuming that the cannonball lands at a point where the height is 0, approximately how many feet farther did the cannonball travel horizontally than vertically upward? (1 yard = 3 feet)

A. 335

B.425

C.935

D. 1,390