Quadratic Functions and Equations Quiz

A quadratic function with a maximum must open downward. A parabola that opens upward cannot have a maximum. Therefore, the statement “The graph of a quadratic function that has a maximum opens upward” is never true. The other statements can occur under valid conditions.

Solve x² − 16 = 0 by setting x² = 16. Taking square roots gives x = ±4. These are the solutions and represent x-intercepts. The other options involve incorrect square roots of 16.

A quadratic function has a maximum only if it opens downward. The function y = −x² + 4 opens downward due to the negative coefficient. Therefore, it has a maximum value at its vertex. The other functions open upward and have minimum values instead.

A quadratic function has a minimum when it opens upward. The function y = x² + 5 opens upward because the coefficient of x² is positive. Therefore, its vertex represents the minimum value. The remaining options open downward and have maximum values instead.

The equation y = (x − 3)² − 5 is in vertex form y = a(x − h)² + k. The vertex is (3, −5). Since the coefficient of the squared term is positive, the parabola opens upward, so the vertex represents the minimum value. Therefore, the function has a minimum value of −5, which is the lowest point on the graph.

The y-intercept occurs when x = 0. Substituting x = 0 into y = x² − 4x − 1 gives y = −1. This value represents where the graph crosses the y-axis. Since no other value is involved, −1 is the y-intercept. This method applies to all functions by evaluating the function at x = 0 directly.

A quadratic function has the highest power of the variable equal to 2. Among the options, y = x² − 4x satisfies this condition. The other functions are linear, cubic, or rational. Therefore, only y = x² − 4x qualifies as a quadratic function by definition.

To find x-intercepts, set y = 0. For y = x² − 9, solve x² = 9. Taking square roots gives x = ±3. These are the x-values where the graph intersects the x-axis. Since both satisfy the equation, the x-intercepts are −3 and 3.

Solve x² − 4 = 0 by rearranging to x² = 4. Taking square roots gives x = ±2. These are the solutions where the graph intersects the x-axis. Since both values satisfy the equation, the solution set is −2 and 2.

The axis of symmetry lies midway between the zeros. The midpoint of −1 and 6 is (−1 + 6)/2 = 5/2. This equals 2.5. The axis of symmetry always passes through this midpoint for any quadratic function, regardless of orientation. Therefore, x = 2.5 is the correct answer.

The equation y = −(x − 2)² + 4 opens downward because the coefficient is negative. The vertex is (2, 4). Since the parabola opens downward, the vertex represents the maximum value. Therefore, the function has a maximum value of 4. The x-coordinate does not affect whether it is maximum or minimum.

The y-intercept is found by substituting x = 0. For y = x² − 3x + 3, substituting gives y = 3. The other options produce y-intercepts of 0 or negative values. Since the question asks for a y-intercept of 3, only y = x² − 3x + 3 satisfies this condition.

To find zeros of x² − 8x + 12, factor the expression. The factors are (x − 2)(x − 6). Setting each factor equal to zero gives x = 2 and x = 6. Zeros are the x-values where the graph crosses the x-axis. Since both values satisfy the equation, the correct pair of zeros is 2 and 6.

The axis of symmetry of a quadratic equation ax² + bx + c = 0 is given by x = −b/(2a). For x² − 6x + 5, a = 1 and b = −6. Substituting gives x = 6/2 = 3. This vertical line divides the parabola into two equal halves.

Factor x² + 3x − 10 to solve. The factors are (x + 5)(x − 2). Setting each factor equal to zero gives x = −5 and x = 2. These values satisfy the equation and represent the points where the graph crosses the x-axis. Therefore, the correct solutions are −5 and 2.

Factor x² + x − 6 into (x + 3)(x − 2). Setting each factor equal to zero gives x = −3 and x = 2. These values solve the equation and represent the x-intercepts. Therefore, the correct solution pair is −2 and 3 after rearranging order.

If a quadratic has no real roots, its discriminant is negative, meaning the parabola does not intersect the x-axis. If the vertex lies in the second quadrant, the parabola must open upward to avoid crossing the x-axis. Therefore, the correct statement is that the graph opens upward.

A parabola with axis of symmetry x = −2 has vertex form y = (x + 2)² + k. Among the options, y = (x + 2)² matches this structure. The axis of symmetry is determined directly from the value inside the squared bracket.

A quadratic function is defined as one where the highest power of the independent variable is exactly 2. This property is always true. Other statements may or may not be true depending on the equation. Therefore, “The highest power of the independent variable in a quadratic function is 2” is the correct always-true statement.

The vertex of y = x² + 6x + 5 is found using x = −b/(2a). Here, a = 1 and b = 6, so x = −6/2 = −3. Substituting x = −3 into the equation gives y = −4. Therefore, the vertex is (−3, −4). This point represents the turning point of the parabola.

Zeros at 1 and −4 mean the factors are (x − 1)(x + 4). Expanding gives x² + 3x − 4. Therefore, the function y = x² + 3x − 4 has zeros at the required points. This confirms the correct equation using factorization.