Sketching Quadratic Graphs from Roots and the Turning Point
Sketch a parabola from its roots, y-intercept, line of symmetry and turning point, reading the turning point off the completed square form.
What a learner can do afterwards
- Read the turning point straight from (x + p)² + q
- Mark roots, y-intercept and line of symmetry on a sketch
- Use the sketch to say for which x values the curve sits above the axis
1 · Read
You have plotted parabolas and studied factorisation, completing the square and the discriminant. A sketch is an approximate graph. Roots are x-values; their x-intercepts are points (x, 0). The y-intercept and turning point are points; the symmetry axis is a line. These features guide a sketch but do not always determine its width. In y = ax² + bx + c, the y-intercept is (0, c).
The sign of a in y = ax² + bx + c decides the opening direction. Positive a gives a minimum; negative a gives a maximum. This does not depend on whether roots exist.
For a quadratic with two distinct roots r and s, symmetry puts the axis halfway between them: x = (r + s)/2. For roots 4 and 10, the axis is x = 7. The axis is a line, not another plotted point.
At the y-axis, x = 0. Substituting zero removes both x terms from ax² + bx + c, leaving y = c. For example, y = 3x² + 2x + 9 has y-intercept (0, 9).
In y = (x + p)² + q, the square is zero at x = -p. The turning point is (-p, q). For y = (x + 6)² - 8, the turning point is (-6, -8).
For y = ax² + bx + c, with a nonzero, the symmetry axis is x = -b/(2a). For y = 2x² - 12x + 7, this gives x = 12/4 = 3.
Completing the square reveals the minimum y-coordinate. The identity x² + 2kx + c = (x + k)² + c - k² shows that this coordinate is c - k². Keep the x-coordinate and y-coordinate separate.
For y = (x - h)² + k, the turning point's x-coordinate is h because x - h must be zero. For y = (x - 9)² + 1, it is 9.
Find roots by setting y = 0 and factorising. If y = (x + 3)(x + 8), then x + 3 = 0 or x + 8 = 0. The roots are -3 and -8. A root is the value that makes a factor zero, not the number written after x.
Use x = -b/(2a) even when a is negative or not 1. With a = -4 and b = 3, the axis is x = -3/(-8) = 0.375. To two decimal places this is 0.38.
For a downward quadratic, complete the square after taking out the minus sign. y = -x² + 12x - 8 becomes -(x - 6)² + 28. The negative square cannot exceed zero, so the maximum y-value is 28.
Roots can also come from completed-square form. For y = (x - 6)² - 25, set y = 0. Then x - 6 = 5 or -5, so x = 11 or 1. Keep both square-root signs.
A downward-opening quadratic with distinct roots r < s has y > 0 only for r < x < s. At either root, y = 0, so exclude the endpoints. Outside the roots, y < 0. An upward-opening quadratic reverses these signs.
A quadratic ax² + c has no x term. Substituting x and -x gives the same output, so its symmetry axis is x = 0. The formula agrees: b = 0 gives -b/(2a) = 0.
To check a proposed turning point for (x + p)² + q, substitute its proposed x-coordinate into the bracket. The bracket must be zero. The y-coordinate is then q. A positive p therefore gives a negative turning-point x-coordinate.
To find a minimum from standard form, halve the x coefficient and correct the constant. x² + 12x + 8 = (x + 6)² - 36 + 8 = (x + 6)² - 28. The minimum y-coordinate is -28.
Factor signs explain sign intervals. For r < x < s, x - r is positive and x - s is negative. Their product is negative. Multiplying by negative a makes y positive. At r or s the product is zero, not positive.
If y = (x - h)² + k with k > 0, every y-value is at least k. The graph never reaches y = 0. Its positive minimum proves there are no real x-axis crossings.
A repeated root gives one contact point. For y = (x + 7)², y = 0 only at x = -7. The square is positive on both sides, so the parabola touches the axis at (-7, 0) without crossing it.
To find the upper endpoint of a positive interval, first find both roots. y = -x² + 12x - 20 = -(x - 2)(x - 10) has roots 2 and 10. It is above the axis for 2 < x < 10; the upper boundary is 10, excluded.
For a scaled quadratic, take out its leading coefficient before completing the square. -3x² + 18x - 2 = -3(x² - 6x) - 2 = -3(x - 3)² + 25. Its maximum y-coordinate is 25.
In y = (x - a)² + b, the symmetry axis is x = a. If two distinct roots are r and s, their midpoint is also on that axis. Therefore a = (r + s)/2. This connects root information with completed-square form.
A negative discriminant means no real roots. If the leading coefficient is also negative, the entire quadratic lies below the axis. For -x² - 4x - 7, D = 16 - 28 = -12. Its completed form is -(x + 2)² - 3, confirming y <= -3.
The midpoint-of-roots rule applies to a quadratic graph with two distinct real roots. Add the roots, then divide by 2. Negative values are included in the sum with their signs. Roots of an arbitrary curve do not establish a symmetry axis.
Reflection across x = h keeps y unchanged and sends x to 2h - x. Across x = 4, a point with x = 1 has partner x = 7. The two horizontal distances from the axis are equal.
Coincident landmarks need an extra point to show width. y = 2x² and y = 3x² share the root x = 0. Their x-intercept, y-intercept and turning point coincide at (0, 0), and both have axis x = 0. At x = 1 their outputs differ. Comparing a point away from the axis distinguishes them.
Count distinct coordinates, not landmark names. In y = (x + 8)², the root is x = -8. Its x-intercept and turning point coincide at (-8, 0). The y-intercept is (0, 64). That gives two distinct points; the symmetry line is not a point.
Mark roots, y-intercept, symmetry line and turning point, then sketch. In (x + p)² + q, the turning point is (-p, q). Use roots and opening direction to locate where y > 0.
Additional questions and readings written for Marble. Videos credit Khan Academy.
2 · Watch
3 · Play
Move a, b and c, and watch the U flip over, the roots slide, and the turning point follow.
Take it off screen
Where it sits
Where this leads
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