College Algebra Quiz: Intercepts And Symmetry

What this quiz covers

This quiz focuses on Intercepts And Symmetry, giving you a quick way to practice the rules, question types, and explanations that matter most for College Algebra.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

Question 1

1. The function has even symmetry, y-intercept at $-3$, and x-intercepts that can be determined by solving $ax^4 + bx^2 + c = 0$
2. The function has odd symmetry, y-intercept at $-3$, and exactly two x-intercepts due to the symmetry constraint
3. The function has even symmetry, y-intercept at $-3$, and exactly four x-intercepts arranged symmetrically about the origin
4. The function has even symmetry, y-intercept at $-3$, and the number of x-intercepts depends on the specific values of $a$ and $b$ (correct answer)

Explanation: From the given points, we can determine that $p(0) = c = -3$, so the y-intercept is at $-3$. Since $p(x) = ax^4 + bx^2 + c$ contains only even powers of $x$, the function is even: $p(-x) = a(-x)^4 + b(-x)^2 + c = ax^4 + bx^2 + c = p(x)$. This is confirmed by the fact that $p(-2) = p(2) = 5$. To find the specific function, we use $p(2) = 5$: $a(2)^4 + b(2)^2 + c = 5$, so $16a + 4b - 3 = 5$, giving $16a + 4b = 8$, or $4a + b = 2$. This gives us one equation with two unknowns, so we cannot determine unique values for $a$ and $b$. For x-intercepts, we solve $ax^4 + bx^2 + c = 0$, or $ax^4 + bx^2 - 3 = 0$. Let $u = x^2$, then $au^2 + bu - 3 = 0$. Using the quadratic formula: $u = \frac{-b ± \sqrt{b^2 + 12a}}{2a}$. For real solutions, we need $b^2 + 12a ≥ 0$. Since $b = 2 - 4a$, we have $(2-4a)^2 + 12a ≥ 0$, which gives $4 - 16a + 16a^2 + 12a ≥ 0$, or $16a^2 - 4a + 4 ≥ 0$. This simplifies to $4a^2 - a + 1 ≥ 0$. The discriminant of this quadratic in $a$ is $1 - 16 = -15 < 0$, so this inequality is always satisfied. However, we also need the solutions for $u$ to be positive (since $u = x^2$). The number of x-intercepts depends on how many positive solutions $u$ has, which depends on the specific values of $a$ and $b$. We could have 0, 2, or 4 x-intercepts. Choice A is incomplete about the number of intercepts. Choice B incorrectly states odd symmetry. Choice C incorrectly assumes exactly four x-intercepts.

Question 2

1. Exactly two x-intercepts at $x = ±2$ (correct answer)
2. Exactly four x-intercepts, because degree 4 polynomials with even symmetry must have symmetric intercept pairs
3. Either two or four x-intercepts, depending on the specific polynomial coefficients
4. Cannot be determined without additional information about the polynomial's coefficients

Explanation: Since $f(x)$ has degree 4, leading coefficient 1, and even symmetry ($f(x) = f(-x)$), it must have the form $f(x) = x^4 + ax^2 + b$ for some constants $a$ and $b$. Using the given conditions: $f(0) = b = -8$ and $f(2) = 16 + 4a - 8 = 0$, so $4a = -8$ and $a = -2$. Therefore, $f(x) = x^4 - 2x^2 - 8$. To find x-intercepts, solve $x^4 - 2x^2 - 8 = 0$. Substituting $u = x^2$: $u^2 - 2u - 8 = 0$. Using the quadratic formula: $u = \frac{2 ± 6}{2}$, giving $u = 4$ or $u = -2$. Since $u = x^2 ≥ 0$, only $u = 4$ is valid, so $x^2 = 4$ and $x = ±2$. We can verify: $f(x) = (x^2 - 4)(x^2 + 2) = (x-2)(x+2)(x^2+2)$. Since $x^2 + 2 > 0$ for all real $x$, the only real zeros are $x = ±2$. Choice B is incorrect because not all degree 4 even polynomials have four real roots. Choices C and D are incorrect because the specific conditions given uniquely determine the polynomial and its intercepts.

Question 3

1. The x-intercepts are wrong; they should be at $x = -4, 4$ since those make the numerator zero
2. The symmetry is wrong; the function is actually odd, not even, due to the rational form
3. The x-intercepts are wrong; $x = ±2$ are not in the domain of the function (correct answer)
4. Both parts are wrong; the x-intercepts are at $x = ±4$ and the function is odd

Explanation: First, let's simplify the function: $f(x) = \frac{x^4 - 16}{x^2 - 4} = \frac{(x^2)^2 - 4^2}{x^2 - 4} = \frac{(x^2 - 4)(x^2 + 4)}{x^2 - 4}$. For $x ≠ ±2$, this simplifies to $f(x) = x^2 + 4$. However, the original function is undefined when $x^2 - 4 = 0$, which occurs at $x = ±2$. Therefore, the domain excludes $x = ±2$. To find x-intercepts, we set the numerator equal to zero: $x^4 - 16 = 0$, so $x^4 = 16$, giving $x = ±2$. However, since $x = ±2$ are not in the domain (they make the denominator zero), there are actually no x-intercepts. The student's claim that x-intercepts are at $x = ±2$ is incorrect because these points are not in the domain. For the simplified form $f(x) = x^2 + 4$ (valid for $x ≠ ±2$), this is indeed an even function since $f(-x) = (-x)^2 + 4 = x^2 + 4 = f(x)$, and $x^2 + 4 > 0$ for all real $x$, confirming no x-intercepts exist. Choice A is wrong because $x = ±4$ don't make the numerator zero. Choice B is wrong because the function is even. Choice D combines two incorrect statements.

Question 4

1. The x-intercepts are located at $x = ±k$ for some positive value $k$, and the y-intercept is at $(0, 6)$ (correct answer)
2. One x-intercept is at $x = 0$ and the other is at some $x = k > 0$, with y-intercept at $(0, 6)$
3. The x-intercepts could be at any two points $x = a$ and $x = b$ where $a ≠ -b$, with y-intercept at $(0, 6)$
4. The x-intercepts are symmetric about the y-axis, but their exact locations cannot be determined from the given information

Explanation: Since $f(x)$ has even symmetry, we know that $f(-x) = f(x)$ for all $x$ in the domain. This means the graph is symmetric about the y-axis. If the function has exactly two x-intercepts and even symmetry, these intercepts must be symmetrically placed about the y-axis. Since $f(0) = 6 ≠ 0$, the point $(0, 0)$ is not an x-intercept. Therefore, the two x-intercepts must be at $x = k$ and $x = -k$ for some positive value $k$. The y-intercept occurs where the graph crosses the y-axis, which is at $x = 0$. Since $f(0) = 6$, the y-intercept is at $(0, 6)$. Choice B is incorrect because if $x = 0$ were an x-intercept, then $f(0) = 0$, but we're told $f(0) = 6$. Choice C is incorrect because even symmetry requires that if $x = a$ is an intercept, then $x = -a$ must also be an intercept. Choice D is correct about the symmetry but incorrectly states that locations cannot be determined - they must be at $x = ±k$ for some positive $k$.