The quadratic function and its graph — Mathematics

In this lesson you will learn

Draw the graphs of y = ax² and y = a(x − m)² + n and explain how a parabola is shifted
Find the vertex, the axis of symmetry and the intercepts of y = ax² + bx + c and sketch it step by step
Read the number of roots, the sign of the function and the signs of a, b and c from a graph
Find the greatest or least value of a quadratic function and write the formula of a parabola from given points

A basketball flying towards the hoop, the jet of a fountain rising and falling back — if we ignore air resistance, all these paths are parabolas. In the lesson “Quadratic equations” you learned to solve ax² + bx + c = 0. Now we look at ax² + bx + c as a function whose value changes as x changes. From its graph you can read off the highest point of a ball, the largest area a fence can enclose, and the roots of the equation.

In the lesson “Linear functions and their graphs” we saw that the graph of y = kx + b is a straight line. The graph of a quadratic function is curved, but it too is completely fixed by a few key points. In this lesson you will learn to find them; later, in the lesson “Quadratic inequalities and the interval method”, the same sketch will help you solve inequalities.

The function y = ax²

Definition

Quadratic function

A function given by y = ax² + bx + c, where a, b, c are numbers and a ≠ 0. Its domain is all real numbers, and its graph is a curve called a parabola.

The simplest case is y = x². Look at the table: x and −x give the same y, because (−x)² = x². So the graph is symmetric about the y-axis. The lowest point is (0, 0): this is the vertex of the parabola. Moving away from the vertex, y grows faster and faster: from x = 1 to 2 it rises by 3, from 2 to 3 by 5. That is why a parabola is flat near the vertex and steep along its arms.

x	−3	−2	−1	1	2	3
y = x²	9	4	1	1	4	9
y = 2x²	18	8	2	2	8	18
y = ½x²	4.5	2	0.5	0.5	2	4.5
y = −x²	−9	−4	−1	−1	−4	−9

For the same x, 2x² is twice as large, ½x² is half as large, and −x² has the opposite sign. Every row is symmetric about 0.

y = ax²

where:

sign of athe direction of the arms: a > 0 — up, a < 0 — down
|a|how wide the parabola is: |a| > 1 — narrow, 0 < |a| < 1 — wide

If a > 0 the vertex is the lowest point, if a < 0 it is the highest. The vertex is (0, 0) and the axis of symmetry is the y-axis. The graph of y = −ax² is the mirror image of y = ax² in the x-axis.

Three questions about y = ax²

1) Does the point A(−3, 18) lie on the parabola y = 2x²?
2) The parabola y = ax² passes through B(2, −12). Find a and say which way the arms point.
3) Find the value of y = ½x² at x = 4 and at x = −4.

Show solution

1) 2 · (−3)² = 2 · 9 = 18 — the equality holds, so A lies on the parabola.
2) −12 = a · 2² ⇒ 4a = −12 ⇒ a = −3. Since a < 0, the arms point down.
3) ½ · 4² = ½ · 16 = 8; x = −4 also gives 8 — by symmetry y(−4) = y(4).

Shifting the parabola: y = a(x − m)² + n

Every value of y = x² + 3 is 3 more than the value of y = x², so the whole graph moves 3 units up. The function y = (x − 2)² takes at x = 2 the value (zero) that y = x² takes at x = 0: every value arrives 2 units “later”, so the graph moves 2 units to the right. Note: a “−” inside the bracket moves it right, a “+” moves it left.

y = a(x − m)² + n

where:

mhorizontal shift: m > 0 — to the right, m < 0 — to the left
nvertical shift: n > 0 — up, n < 0 — down
aas in y = ax²: the direction and the width

The vertex is (m, n) and the axis of symmetry is the line x = m. The graph is y = ax² moved m units horizontally and n units vertically; its shape does not change.

Parabolas given in vertex form

1) Find the vertex of y = (x + 2)² − 1 and say how it is obtained from y = x².
2) Find the vertex of y = −2(x − 1)² + 8 and the points where it crosses the x-axis.
3) The parabola y = 3x² was moved 4 units to the left and 5 units down. Write the formula of the new parabola.

Show solution

1) (x + 2)² = (x − (−2))², so m = −2 and n = −1. The vertex is (−2, −1); the graph is y = x² moved 2 units left and 1 unit down.
2) The vertex is (1, 8); a = −2 < 0, so the arms point down and 8 is the greatest value.
y = 0: −2(x − 1)² + 8 = 0 ⇒ (x − 1)² = 4 ⇒ x − 1 = ±2 ⇒ x = 3 and x = −1.
3) 4 to the left gives x + 4 in the bracket, 5 down gives −5 at the end: y = 3(x + 4)² − 5, vertex (−4, −5).

Interactive

Loading simulation…

Move the m and n sliders: the parabola slides without changing its shape, and its vertex stays at (m, n). Change a: its sign flips the arms, and its size makes the parabola narrower or wider.

y = ax² + bx + c: vertex, axis of symmetry and intercepts

Every quadratic function can be brought to vertex form by completing the square (the method is in the lesson “Quadratic equations”): ax² + bx + c = a(x + b/(2a))² − (b² − 4ac)/(4a). So m = −b/(2a) and n = −D/(4a). This means the graph of y = ax² + bx + c is simply a shifted copy of y = ax².

x₀ = −b/(2a), y₀ = f(x₀) = −D/(4a)x₀ = −b/(2a), y₀ = f(x₀) = −D/(4a)

where:

x₀the x-coordinate of the vertex; the axis of symmetry is the line x = x₀
y₀the y-coordinate of the vertex — the least (a > 0) or greatest (a < 0) value of the function
Dthe discriminant b² − 4ac

In practice it is easier to find y₀ by substituting x₀ into the function than by using the formula.

Finding the vertex

Find the vertex of the parabola: 1) y = x² − 6x + 5; 2) y = −2x² + 8x − 3; 3) y = 3x² + 12x.

Show solution

1) x₀ = 6/2 = 3, y₀ = 9 − 18 + 5 = −4. The vertex is (3, −4); vertex form: y = (x − 3)² − 4.
2) x₀ = −8/(2 · (−2)) = 2, y₀ = −8 + 16 − 3 = 5. The vertex is (2, 5). Check with the formula: D = 64 − 24 = 40, −D/(4a) = −40/(−8) = 5.
3) x₀ = −12/6 = −2, y₀ = 12 − 24 = −12. The vertex is (−2, −12); since c = 0, the parabola passes through the origin.

The y-intercept: put x = 0 and you get y = c — the point (0, c). It is the one point you get with no calculation. The x-intercepts: y = 0, that is, the equation ax² + bx + c = 0. If D > 0 the parabola crosses the x-axis at two points; if D = 0 it touches the axis at its vertex; if D < 0 it has no common point with the x-axis. The x-coordinates of these points are the zeros of the function.

Sketching the graph step by step and reading it

1
Direction of the arms
Look at the sign of a: a > 0 — up, a < 0 — down.
2
Vertex and axis
Compute x₀ = −b/(2a) and y₀ = f(x₀); mark the vertex and draw the axis x = x₀ as a dashed line.
3
The y-intercept
Mark (0, c) and its mirror image (2x₀, c).
4
Zeros
If D ≥ 0, solve ax² + bx + c = 0 and mark the roots on the x-axis.
5
Extra points
If needed, compute one or two points further from the vertex and reflect them too.
6
Draw a smooth curve
Join the points with a smooth curve that turns round at the vertex without a sharp corner, and label the graph.

The graph of y = x² − 2x − 3 and what it tells us

Sketch the graph of y = x² − 2x − 3. Use it to find the range, where the function increases and decreases, and for which x it is positive or negative.

Show solution

1) a = 1 > 0 — the arms point up.
2) x₀ = 2/2 = 1, y₀ = 1 − 2 − 3 = −4. Vertex (1, −4), axis x = 1.
3) y-intercept: (0, −3), its mirror image (2, −3).
4) x² − 2x − 3 = 0 ⇒ x = 3, x = −1 (a − b + c = 0).
5) x = 4: y = 5; the mirror point x = −2: y = 5.
Reading the graph: the range is y ≥ −4 (least value −4); the function decreases on (−∞, 1] and increases on [1, +∞); y < 0 for −1 < x < 3 (the graph is below the x-axis) and y > 0 for x < −1 or x > 3.

Interactive

Loading simulation…

It starts as y = x² − 2x − 3. Change c: the parabola slides up and down. Change b: the point (0, c) stays put while the vertex moves sideways and up or down (it travels along the parabola y = c − ax²). Change the sign of a: the arms flip.

You can also read the signs of the coefficients from the graph. a shows the direction of the arms. c is the y-coordinate of the point where the graph crosses the y-axis. b, together with a, fixes the position of the vertex: since x₀ = −b/(2a), the vertex is to the left of the y-axis when a and b have the same sign, to the right when their signs differ, and on the y-axis when b = 0. The number of common points with the x-axis gives the sign of D.

What you see on the graph	Conclusion
Arms point up / down	a > 0 / a < 0
Narrow / wide parabola	\|a\| large / small
Crosses the y-axis above / below the origin	c > 0 / c < 0
Passes through the origin	c = 0
Vertex to the left of the y-axis	x₀ < 0: a and b have the same sign
Vertex to the right of the y-axis	x₀ > 0: a and b have different signs
Vertex on the y-axis	b = 0
2 / 1 / 0 common points with the x-axis	D > 0 / D = 0 / D < 0

Reading the signs from the graph

The parabola y = ax² + bx + c opens downwards, crosses the y-axis below the origin, has its vertex to the right of the y-axis and crosses the x-axis at two points. Find the signs of a, b, c and D and give an example of such a function.

Show solution

Arms down ⇒ a < 0. The y-intercept is below the origin ⇒ c < 0.
Vertex on the right: x₀ = −b/(2a) > 0 ⇒ a and b have different signs ⇒ b > 0.
Two crossing points ⇒ D > 0.
Example: y = −x² + 4x − 3: a = −1, b = 4, c = −3, D = 16 − 12 = 4 > 0; vertex (2, 1), zeros 1 and 3.

Greatest and least values. Finding the formula from points

a > 0: min y = y₀; a < 0: max y = y₀; y₀ = f(−b/(2a))a > 0: min y = y₀; a < 0: max y = y₀; y₀ = f(−b/(2a))

where:

min ythe least value of the function
max ythe greatest value of the function
−b/(2a)−b/(2a)the point where this value is reached — the x-coordinate of the vertex

A parabola that opens upwards has its lowest point at the vertex and no upper bound; one that opens downwards is the other way round. The range is y ≥ y₀ if a > 0 and y ≤ y₀ if a < 0.

A thrown ball

Murad throws a ball upwards from a height of 1.8 m. After t seconds the ball is at height h = 1.8 + 12t − 5t² metres. What is its greatest height, and when is it back at 1.8 m?

Show solution

a = −5 < 0, so the arms point down and the vertex is the highest point.
t₀ = −12/(2 · (−5)) = 1.2 s; h(1.2) = 1.8 + 14.4 − 7.2 = 9 m.
Height 1.8 m: 12t − 5t² = 0 ⇒ t(12 − 5t) = 0 ⇒ t = 0 (the throw) or t = 2.4 s.
Notice the symmetry: 2.4 = 2 · 1.2 — the ball rises for 1.2 s and falls for 1.2 s.

The largest area and the largest product

1) Leyla has 24 m of fencing. What size of rectangular plot should she fence to get the largest area?
2) What if one side of the plot is a wall and the 20 m of fencing goes only along the other three sides?
3) What is the largest possible product of two numbers whose sum is 10?

Show solution

1) Let one side be x m; then the neighbouring side is 12 − x m (half the perimeter is 12). S = x(12 − x) = −x² + 12x.
a = −1 < 0, x₀ = −12/(−2) = 6 ⇒ 6 m × 6 m, S = 36 m² — a square.
2) The two sides perpendicular to the wall are x m, the side parallel to it is 20 − 2x m (0 < x < 10). S = x(20 − 2x) = −2x² + 20x.
x₀ = −20/(−4) = 5 ⇒ 5 m × 10 m, S = 50 m².
3) The numbers are x and 10 − x: P = −x² + 10x, x₀ = 5 ⇒ the largest product is 5 · 5 = 25.
The general method for any function, using the derivative, is in the lesson “Greatest and least values on a closed interval. Optimisation problems”.

To find the formula of a parabola you usually need three conditions, because the formula has three coefficients. Choose one of three forms depending on the data: if the vertex is known, y = a(x − m)² + n; if the zeros are known, y = a(x − x₁)(x − x₂); if only three points are known, y = ax² + bx + c and a system of three equations.

y = a(x − x₁)(x − x₂)

where:

x₁, x₂the zeros of the function — where the parabola crosses the x-axis
afound from one more point

This is the factored form of a quadratic function (the formula ax² + bx + c = a(x − x₁)(x − x₂) from the lesson “Quadratic equations”); it exists when D ≥ 0. The vertex is halfway between the roots: x₀ = (x₁ + x₂)/2.

Finding the formula

Find the formula y = ax² + bx + c of the parabola that:
1) has its vertex at (2, −3) and passes through (0, 5);
2) crosses the x-axis at x = −1 and x = 3 and passes through (0, −6);
3) passes through (0, 1), (1, 2) and (−1, 6).

Show solution

1) y = a(x − 2)² − 3; x = 0: 5 = 4a − 3 ⇒ a = 2.
y = 2(x − 2)² − 3 = 2x² − 8x + 5.
2) y = a(x + 1)(x − 3); x = 0: −6 = a · 1 · (−3) ⇒ a = 2.
y = 2(x + 1)(x − 3) = 2x² − 4x − 6.
3) x = 0 ⇒ c = 1. x = 1: a + b + 1 = 2; x = −1: a − b + 1 = 6.
a + b = 1 and a − b = 5 ⇒ a = 3, b = −2. y = 3x² − 2x + 1.
Check: y(−1) = 3 + 2 + 1 = 6 — correct.

Check yourself

1.The x-coordinate of the vertex of y = 2x² − 12x + 1 is x₀ =
2.The least value of y = x² + 4x + 1 is
3.The greatest value of y = −x² + 6x − 5 is
4.The parabola y = 3x² − x − 7 crosses the y-axis at y =
5.The axis of symmetry of y = (x − 1)(x − 9) is x =

Key points

The graph of y = ax² + bx + c (a ≠ 0) is a parabola: a > 0 — arms up, a < 0 — arms down; the larger |a|, the narrower the parabola.
y = a(x − m)² + n is y = ax² shifted m units horizontally and n units vertically; the vertex is (m, n).
Vertex: x₀ = −b/(2a), y₀ = f(x₀) = −D/(4a); the axis of symmetry is x = x₀; if there are roots, x₀ is their midpoint.
The y-intercept is (0, c); the x-intercepts are the roots of ax² + bx + c = 0: D > 0 — two points, D = 0 — the parabola touches the axis, D < 0 — no common points.
If a > 0 the least value, and if a < 0 the greatest value, is y₀, reached at the vertex.
The formula is found from three conditions: via the vertex form, the factored form, or three points.

Check yourself

12 questions. Every correct answer earns XP.

1 / 12

Which way do the arms of y = −3x² + x + 7 point?