Learning Objectives
Read vertex, symmetry and intercepts from useful quadratic forms.
Find a quadratic range from its completed square.
Solve line–parabola intersections and test tangency.
Distinguish a global extremum from an endpoint-restricted one.
Prerequisites: quadratic roots, completing the square and discriminants; coordinates and simultaneous equations. All practice questions in this lesson are original, not official TMUA questions.
What This Lesson Is About
A quadratic graph is a parabola. Its algebraic expression can reveal where it crosses the axes, where it turns, and which output values it reaches. This lesson connects those features to the forms of a quadratic you already know.
We then find where a line and parabola meet, use repeated roots to identify tangency, and determine how restricting the allowed inputs changes the range.
1. Reading a Quadratic Graph from Its Equation
The same quadratic can be written in several equivalent forms. Choose the form that makes the graph feature you need easiest to identify.
1.1 Intercepts, vertex and symmetry from three forms
The expanded form y = ax² + bx + c gives the vertical-axis intercept (0, c). The factored form y = a(x − r)(x − s) gives horizontal-axis intercepts at r and s when these roots are real. The completed-square form y = a(x − h)² + k gives the vertex (h, k) and symmetry line x = h.
1.2 Opening direction and the turning value
For a > 0 the parabola opens upward and k is its minimum. For a < 0 it opens downward and k is its maximum. Equal displacements on either side of h have equal squares and therefore equal heights.
1.3 Finding the range when all real inputs are allowed
When x ranges over all real numbers, the range is y ≥ k for a > 0 and y ≤ k for a < 0. This follows from the nonnegativity of a square and the fact that every nonnegative square value is attainable.
Figure 1. The computed curve passes through all four labelled points. The dashed line x = 2 is the axis of symmetry.
Worked Example: One Graph, Three Descriptions
Question: Describe y = x² − 4x + 3.
Method: Factor to find the horizontal intercepts and complete the square to find the vertex. Use the expanded form for the vertical intercept.
Factor to get (x − 1)(x − 3), giving roots 1 and 3. Complete the square to get (x − 2)² − 1, giving vertex (2, −1), symmetry x = 2, and minimum −1.
The y-intercept is (0, 3). Symmetry gives the matching point (4, 3). These exact points anchor the sketch rather than relying on its appearance.
2. Finding Intersections and Tangency
An intersection belongs to both graphs, so its coordinates must satisfy both equations. Equating their outputs reduces this geometric question to a quadratic equation.
2.1 Solving for both coordinates of each intersection
Where two graphs meet, they share both coordinates. Equate their expressions for y and solve for x, then substitute each x into either original graph to recover y. Check both equations at every resulting point.
2.2 Using the root count to recognise crossings or tangency
A line and a genuine parabola can have two, one or no real intersection points. In the resulting quadratic, a repeated root gives one tangency point; two distinct roots give two crossings.
Worked Example: Intersect a Line and a Parabola
Question: Find intersections of y = x² − 4x + 3 and y = x − 1.
Method: Equate the two expressions for y, solve for x, and substitute each root to obtain the corresponding y-coordinate.
Equating gives x² − 5x + 4 = (x − 1)(x − 4) = 0. Therefore x = 1 or 4.
Using y = x − 1 gives (1, 0) and (4, 3). Substituting into the parabola gives 1 − 4 + 3 = 0 and 16 − 16 + 3 = 3.
Worked Example: A Tangent without Differentiation
Question: For which c is y = 2x + c tangent to y = x²?
Method: Form the intersection equation and require a repeated root. Substitute that root to find the contact point.
Intersections satisfy x² − 2x − c = 0. One repeated root requires Δ = 4 + 4c = 0, hence c = −1.
The equation then becomes (x − 1)² = 0, so the contact point is (1, 1). Also x² − (2x − 1) = (x − 1)² ≥ 0, confirming the line never lies above the parabola.
3. Finding the Range on a Restricted Domain
The vertex gives the turning value of the whole parabola, but it may not be available on a specified input interval. The range must use only points within that domain.
3.1 Comparing distances from the vertex
If only an interval of x values is allowed, first check whether it contains the vertex. Compare the squared displacements from the vertex: the smallest gives the extremum closest to the turning point, and the largest gives the opposite extreme on a closed interval.
3.2 Checking whether endpoint values are attained
For an open endpoint, a limiting bound may not be attained. State the difference between “greater than” and “greater than or equal to” in a range; do not assume every bound is a maximum or minimum.
Worked Example: An Interval to the Right of the Vertex
Question: Find the range of y = (x − 2)² − 1 for 3 ≤ x ≤ 5.
Method: Check whether the domain contains the vertex, then compare the smallest and largest squared displacements on the interval.
Here 1 ≤ x − 2 ≤ 3, so 1 ≤ (x − 2)² ≤ 9. Subtracting 1 gives 0 ≤ y ≤ 8.
The endpoints are attained at x = 3 and x = 5. The global minimum −1 is not available because x = 2 is outside the domain.
Common Mistakes
The vertex x-coordinate in (x − h)² is h, not −h.
A y-intercept is found with x = 0, while a root is found with y = 0.
An intersection answer needs paired coordinates, not only x values.
A restricted domain may exclude the global vertex or an endpoint bound.
4. Practice
Use exact algebra to identify the key graph features. Give intersections as coordinate pairs, and check whether the vertex and endpoints belong to the stated domain before giving a range.
1. Find vertex, symmetry line and range of y = 2(x + 1)² − 8 for real x.
2. Find both axis intercepts of y = x² + x − 6.
3. Find intersections of y = x² and y = 3x + 4.
4. Find c for tangency of y = 4x + c to y = x².
5. Find the range of y = −(x − 1)² + 4 for 0 ≤ x ≤ 3.
6. For which t does the horizontal line y = t meet y = (x + 2)² + 5 twice?
Worked Solutions
1. Reading vertex, symmetry and range
The vertex is (−1, −8), symmetry line x = −1, and range y ≥ −8. The square is zero at −1 and otherwise nonnegative.
2. Finding intercepts on both axes
At x = 0 the point is (0, −6). Factoring (x + 3)(x − 2) gives horizontal intercepts (−3, 0) and (2, 0).
3. Recovering both intersection coordinates
x² − 3x − 4 = (x − 4)(x + 1) = 0, giving (4, 16) and (−1, 1). Both satisfy y = 3x + 4.
4. Finding a tangent using a repeated root
x² − 4x − c = 0 has Δ = 16 + 4c. Thus c = −4, and the equation is (x − 2)² = 0, giving contact (2, 4).
5. Finding a range that includes the vertex
The vertex x = 1 is included, giving maximum 4. The largest distance from 1 is 2 at x = 3, giving minimum 0. Thus 0 ≤ y ≤ 4. At x = 0 the value 3 lies inside this range.
6. Counting intersections with a horizontal line
(x + 2)² = t − 5 has two distinct solutions exactly when t > 5. Equality gives tangency and smaller t gives no intersection.
5. Summary
- Use expanded, factored and completed-square forms for their different information.
- Solve intersections algebraically and recover both coordinates.
- A repeated intersection root identifies tangency to a parabola.
- Restrict the range using the actual allowed x values.