Intercepts and Symmetry - College Algebra

1. For the function $f(x) = \frac{x^4 - 16}{x^2 - 4}$, a student claims that the x-intercepts are at $x = -2, 2$ and the function is even. Which part of this analysis is incorrect?
   Answer: The x-intercepts are wrong; $x = ±2$ are not in the domain of the function 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.
2. A polynomial function $f(x)$ has degree 4, leading coefficient 1, and satisfies $f(x) = f(-x)$ for all $x$. If $f(0) = -8$ and $f(2) = 0$, how many x-intercepts does $f(x)$ have?
   Answer: Exactly two x-intercepts at $x = ±2$ 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.
3. The function $p(x) = ax^4 + bx^2 + c$ passes through the points $(-2, 5)$, $(0, -3)$, and $(2, 5)$. Based on this information, what can be determined about the intercepts and symmetry of $p(x)$?
   Answer: The function has even symmetry, y-intercept at $-3$, and the number of x-intercepts depends on the specific values of $a$ and $b$ 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.
4. A function $f(x)$ has the following properties: it has exactly two x-intercepts, it has even symmetry, and $f(0) = 6$. Which of the following statements must be true?
   Answer: The x-intercepts are located at $x = ±k$ for some positive value $k$, and the y-intercept is at $(0, 6)$ 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 $k$.

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