## Engage NY Eureka Math Algebra 2 Module 1 End of Module Assessment Answer Key

### Eureka Math Algebra 2 Module 1 End of Module Assessment Answer Key

Question 1.

A parabola is defined as the set of points in the plane that are equidistant from a fixed point (called the focus of the parabola) and a fixed line (called the directrix of the parabola).

Consider the parabola with focus point (1, 1) and directrix the horizontal line y = – 3.

a. What are the coordinates of the vertex of the parabola?

Answer:

(1, – 1)

b. Plot the focus and draw the directrix on the graph below. Then draw a rough sketch of the parabola.

Answer:

c. Find the equation of the parabola with this focus and directrix.

Answer:

A point (x, y) on the parabola is equidistant from the directrix and the focus:

d. What is the y-intercept of this parabola?

Answer:

At the y-intercept x = 0, so y = \(\frac{1}{8}\)(-1)^{2} – 1 = \(\frac{-7}{8}\)

\(\frac{-7}{8}\) is the y intercept.

e. Demonstrate that your answer from part (d) is correct by showing that the y-intercept you identified is indeed equidistant from the focus and the directrix.

Answer:

f. Is the parabola in this question (with focus point (1, 1) and directrix y = – 3) congruent to a parabola with focus (2, 3) and directrix y = – 1? Explain.

Answer:

g. Is the parabola in this question (with focus point (1, 1) and directrix y = – 3) congruent to the parabola with equation given by y = x^{2}? Explain.

Answer:

No, y = x^{2} and y = \(\frac{1}{8}\)(x – 1)^{2} + 1 do not have the same leading coefficient so they are not congruent. (y = x^{2} has a leading coefficient 1, and y = \(\frac{1}{8}\)(x – 1)^{2} + 1 had a leading coefficient \(\frac{1}{8}\).

h. Are the two parabolas from part (g) similar? Why or why not?

Answer:

Yes! Because all parabolas are similar.

Question 2.

The graph of the polynomial function f(x) = x^{3} + 4x^{2} + 6x + 4 is shown below.

a. Based on the appearance of the graph, what does the real solution to the equation x^{3} + 4x^{2} + 6x + 4 = 0 appear to be? Jiju does not trust the accuracy of the graph. Prove to her algebraically that your answer is in fact a zero of y = f(x).

Answer:

The real zero appears to be x = -2.

f(-2) = (-2)^{3} + 4(-2)^{2} + 6(-2) + 4

= – 8 + 16 – 12 + 4

= 0

b. Write f as a product of a linear factor and a quadratic factor, each with real number coefficients.

Answer:

c. What is the value of f(10)? Explain how knowing the linear factor of f establishes that f(10) is a multiple of 12.

Answer:

f(10) = (10 + 2) (100 + 20 + 2)

= 12 (122)

= 1220 + 244

= 1464

f(10) is a multiple of 12 because f(x) has a linear factor of x + 2, and x + 2 = 12, when x = 10.

d. Find the two complex number zeros of y = f(x).

Answer:

We need to solve x^{2} + 2x + 2 = 0

x^{2} + 2x + 1 = – 1

(x + 1)^{2} = – 1

x + 1 = Â± 1

x = – 1 Â± l

e. Write f as a product of three linear factors.

Answer:

f(x) = (x + 2) (x – ( – 1 + l)) (x – (- 1 – l)

Question 3.

A line passes through the points (- 1, 0) and P = (0, t) for some real number t and intersects the circle x^{2} + y^{2} = 1 at a point Q different from (- 1, 0).

a. If t = \(\frac{1}{2}\), so that the point P has coordinates (o, \(\frac{1}{2}\)), find the coordinates of the point Q.

Answer:

A Pythagorean triple is a set of three positive integers a, b, and c satisfying a^{2} + b^{2} = c^{2}. For example, setting a = 3, b = 4, and c = 5 gives a Pythagorean triple.

b. Suppose that \(\left(\frac{a}{c}, \frac{b}{c}\right)\) is a point with rational number coordinates lying on the circle x^{2} + y^{2} = 1. Explain why then a, b, and c form a Pythagorean triple.

Answer:

c. Which Pythagorean triple is associated with the point Q = \(\left(\frac{5}{13}, \frac{12}{13}\right)\) on the circle?

Answer:

5, 12, 13

d. If Q = \(\left(\frac{5}{13}, \frac{12}{13}\right)\) what is the value of t so that the point P has coordinates (0, t)?

Answer:

e. Suppose we set x = \(\frac{1-t^{2}}{1+t^{2}}\) and y = \(\frac{2 t}{1+t^{2}}\) for a real number t. Show that (x, y) is then a point on the circle x^{2} + y^{2} = 1.

Answer:

f. Set t = \(\frac{3}{4}\) in the formulas x = \(\frac{1-t^{2}}{1+t^{2}}\) and y = \(\frac{2 t}{1+t^{2}}\). Which point on the circle x^{2} + y^{2} = 1 does this give? What is the associated Pythagorean triple?

Answer:

g. Suppose t is a value greater than 1, P = (0, t), and Q is the point in the second quadrant (different from (- 1,0)) at which the line through (- 1,0) and P intersects the circle x^{2} + y^{2} = 1. Find the coordinates of the point Q in terms of t.

Answer:

Question 4.

a. Write a system of two equations in two variables where one equation is quadratic and the other is linear such that the system has no solution. Explain, using graphs, algebra, and/or words, why the system has no solution.

Answer:

y = x^{2}

y = – 1

From the graph, these two curves do not intersect, and so there is no solution to this system of equations.

b. Prove that x = \(\sqrt{-5 x-6}\) has no solution.

Answer:

c. Does the following system of equations have a solution? If so, find one. If not, explain why not.

2x + y + z = 4

x – y + 3z = – 2

– x + y + z = – 2

Answer:

2x + y + z = 4Â Â Â Â Â â†’Â Â Â Â 1

x – y + 3z = – 2Â Â Â Â â†’ Â Â Â 2

– x + y + z = – 2 Â Â Â â†’ Â Â Â 3