Questions: Graphing Quadratic Functions

Question 1

A student begins graphing y = −3x² + 12x − 5 and expects a U-shaped curve opening upward. What will actually appear?

• A. A U-shaped curve opening upward, since the coefficient of x is positive
• B. A U-shaped curve opening upward, since the equation has three terms
• C. An inverted U-shape opening downward, because the leading coefficient a = −3 is negative
• D. A straight line, because the negative sign cancels the squaring effect

Answer: (C)

The sign of the leading coefficient a determines whether the parabola opens upward (a > 0, a 'smile') or downward (a < 0, a 'frown'). Here a = −3, so the parabola opens downward and has a maximum point. The common misconception is to focus on the positive middle term (+12x) and expect upward opening — but only the sign of the coefficient on x² determines direction. A negative a always produces a downward-opening parabola.

Question 2

For y = 2x² + 8x + 3, a student calculates the vertex x-coordinate as x = b/(2a) = 8/(4) = 2. What error was made?

• A. The student used the wrong value for b — it should be the coefficient of x², not x
• B. The student forgot the negative sign: the correct formula is x = −b/(2a) = −8/4 = −2
• C. The formula for the vertex is x = −b/a, not −b/(2a)
• D. The student calculated correctly; the vertex x-coordinate is 2

Answer: (B)

The vertex formula is x = −b/(2a), with a critical negative sign. The student computed b/(2a) instead, getting the wrong sign. For this function, the correct vertex x-coordinate is −8/(2·2) = −8/4 = −2, not +2. Forgetting the negative sign in the vertex formula is one of the most common errors in graphing quadratics. The symmetry of a parabola means that using +2 instead of −2 produces a point on the curve, but not the vertex.

Question 3

Most parabolas open upward because the x² term generally represents a positive squared value.

• T. True
• F. False

Answer: False

This confuses the sign of x² (which is always non-negative as a mathematical expression) with the sign of the coefficient a. The coefficient a multiplies x², and if a < 0, the product ax² is negative for any nonzero x. This makes the parabola open downward. For example, y = −x² produces y ≤ 0 for all x, giving a downward-opening parabola with a maximum at the origin. The key is whether a is positive or negative, not whether x² is positive.

Question 4

A quadratic function whose discriminant is negative has no graph — since it has no real x-intercepts, the parabola does not exist.

• T. True
• F. False

Answer: False

A parabola with no real x-intercepts still has a complete graph — it simply floats entirely above the x-axis (if a > 0) or below it (if a < 0) without ever touching zero. For example, y = x² + 1 is a perfectly valid upward-opening parabola with vertex at (0, 1) that never crosses the x-axis. The x-intercepts are the solutions to the equation set equal to zero; their absence means no real roots, not no graph. This is an important distinction: every quadratic function has a parabolic graph.

Question 5

Explain how the axis of symmetry can be used to plot a parabola efficiently. What is the relationship between points on either side of the axis?

Model answer: The axis of symmetry is the vertical line x = −b/(2a) that passes through the vertex, dividing the parabola into two mirror-image halves. Any point on the parabola has a corresponding point at equal distance on the other side of the axis with the same y-value. This means once you find the vertex and plot a few points on one side, you can immediately place their mirror images on the other side without additional calculation. For example, if the axis is x = 2 and you find the point (0, 5) on the parabola, then (4, 5) is also on the parabola by symmetry.

This symmetry property follows from the fact that squaring is an even function: (−h)² = h². So the y-value depends only on how far x is from the vertex's x-coordinate, not on which direction. Exploiting this symmetry is much more efficient than plugging in every x-value independently. In practice: find the vertex, find the y-intercept, reflect the y-intercept across the axis, add one or two more calculated points and their reflections — you now have enough to sketch an accurate parabola.