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

## 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?
(1, – 1)

b. Plot the focus and draw the directrix on the graph below. Then draw a rough sketch of the parabola.  c. Find the equation of the parabola with this focus and directrix.
A point (x, y) on the parabola is equidistant from the directrix and the focus: d. What is the y-intercept of this parabola?
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. 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. 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 = x2? Explain.
No, y = x2 and y = $$\frac{1}{8}$$(x – 1)2 + 1 do not have the same leading coefficient so they are not congruent. (y = x2 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?
Yes! Because all parabolas are similar.

Question 2.
The graph of the polynomial function f(x) = x3 + 4x2 + 6x + 4 is shown below. a. Based on the appearance of the graph, what does the real solution to the equation x3 + 4x2 + 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).
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. 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.
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).
We need to solve x2 + 2x + 2 = 0
x2 + 2x + 1 = – 1
(x + 1)2 = – 1
x + 1 = ± 1
x = – 1 ± l

e. Write f as a product of three linear factors.
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 x2 + y2 = 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. A Pythagorean triple is a set of three positive integers a, b, and c satisfying a2 + b2 = c2. 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 x2 + y2 = 1. Explain why then a, b, and c form a Pythagorean triple. c. Which Pythagorean triple is associated with the point Q = $$\left(\frac{5}{13}, \frac{12}{13}\right)$$ on the circle?
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)? 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 x2 + y2 = 1. 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 x2 + y2 = 1 does this give? What is the associated Pythagorean triple? 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 x2 + y2 = 1. Find the coordinates of the point Q in terms of t. 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.
y = x2
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.  