Sketching from Vertex and Factored Form

Each form hands you its own landmarks.

Vertex form

A quadratic written as y = (x − 2)² + 1 is in vertex form, because its vertex, the turning point of the curve, can be read straight off it. The square (x − 2)² is never below 0, and it is 0 when x = 2. So y is never less than 1, and y = 1 when x = 2. The lowest point of the curve is (2, 1).

The curve is symmetric about the vertical line through its vertex, x = 2. At x = 1 and at x = 3, the square is (−1)² = 1 and 1² = 1, so both give y = 2. Points the same distance either side of x = 2 are at the same height.

Two more facts finish the sketch. Where the curve crosses the y-axis, x = 0, so y = (0 − 2)² + 1 = 5. And since the lowest value of y is 1, which is above 0, the curve never meets the x-axis.

(2, 1)

y = (x − 2)² + 1: the vertex (2, 1) is the lowest point, and the dashed line x = 2 is the axis of symmetry. The curve stays above the x-axis.

A minus in front

Now take y = −(x − 2)² + 4. The square is still never below 0, but the minus sign in front turns it round: −(x − 2)² is never above 0. So y is never more than 4, and y = 4 when x = 2. The vertex (2, 4) is now the highest point, and the curve opens downward.

This curve does meet the x-axis. y = 0 when (x − 2)² = 4, so x − 2 = 2 or x − 2 = −2, which gives x = 4 or x = 0. The curve crosses at 0 and 4, the same distance either side of x = 2.

04(2, 4)

y = −(x − 2)² + 4 opens downward from its highest point, (2, 4), and crosses the x-axis at 0 and 4.

Factored form

A quadratic written as y = (x − 1)(x − 5) is in factored form, and it shows where the curve crosses the x-axis. y = 0 when one of the brackets is 0, so the crossings are at x = 1 and x = 5.

The curve is symmetric, so its turning point is halfway between the crossings: x = (1 + 5) / 2 = 3. Put x = 3 into the equation to find how low it goes: y = (3 − 1)(3 − 5) = 2 × (−2) = −4. The vertex is (3, −4).

Multiplying the brackets out would start with x², whose coefficient is positive, so the curve opens upward. It crosses the y-axis at y = (0 − 1)(0 − 5) = (−1) × (−5) = 5.

15(3, −4)

y = (x − 1)(x − 5) crosses the x-axis at 1 and 5 and turns halfway between them, at (3, −4).

−5−4−3−2−10123456x = −0.5x² − (−2 + 1)x + (−2)(1)= x² + x − 2

sum −1, product −2: x² − (sum)x + product, so the middle coefficient is −−1 and the constant is −2; the axis of symmetry x = −0.5 is half the sum

Put the roots at 2 and 6 and read the expansion

The curve y = (x − r₁)(x − r₂), with its two crossings as the handles. However you move them, the dashed line through the turning point stays halfway between them.

A minus in front of the brackets

y = −(x − 1)(x − 5) is 0 at the same two places, x = 1 and x = 5. Every other value of y has its sign changed, so the curve is turned upside down. At x = 3 it is −(2 × (−2)) = 4, so the turning point (3, 4) is now the highest point, and the curve opens downward.

15(3, 4)

y = −(x − 1)(x − 5) keeps the crossings at 1 and 5 and turns over the top, at (3, 4).

One curve, two forms

Both forms can describe the same curve. y = x² − 6x + 5 factors as (x − 1)(x − 5), and completing the square gives (x − 3)² − 9 + 5 = (x − 3)² − 4. The factored form gives the crossings, 1 and 5. The vertex form gives the vertex, (3, −4). Both agree: 3 is halfway between 1 and 5.

So to sketch a quadratic, write it in whichever form gives the landmark you need: vertex form for the turning point, factored form for the crossings. Add the y-intercept and whether the curve opens upward or downward.

The usual mistakes

Reading the vertex of y = (x − 2)² + 1 as (−2, 1). The square is 0 when x = 2, so the vertex is at x = 2.

Putting the turning point at a crossing. For y = (x − 1)(x − 5), the curve turns at x = 3, halfway between 1 and 5, not at 1 or at 5.

Forgetting what the minus in front does. It turns the curve upside down: the vertex becomes the highest point, and the curve opens downward.

Where a quadratic is above zero

A sketch also shows where a quadratic is positive and where it is negative. y = (x − 1)(x − 5) opens upward, so it is below the x-axis between its crossings and above it outside them. So (x − 1)(x − 5) < 0 when 1 < x < 5, and (x − 1)(x − 5) > 0 when x < 1 or x > 5. Check x = 3: (3 − 1)(3 − 5) = −4, which is negative.

y = −(x − 1)(x − 5) opens downward, so the two parts swap: −(x − 1)(x − 5) > 0 exactly when 1 < x < 5. So a quadratic inequality is solved by finding the crossings, then reading off the sketch which side of the x-axis the curve is on.

Worked example: The Range of Prices for Which a Stall Makes a Profit

Question A drinks stall sells a cup of juice for x dollars. Its profit for a day is P dollars, where P = −5x2 + 60x − 100. (a) For which prices does the stall make a profit? (b) Which price gives the greatest profit, and how much is that profit?

  1. 1.The stall makes a profit when P > 0, so −5x2 + 60x − 100 > 0. Divide both sides by −5. Dividing by a negative number reverses the inequality sign: x2 − 12x + 20 < 0.

    −100−50050100024681012price in dollars, xprofit in dollars, Pa profit: −5x2+ 60x − 100 > 0divide both sides by −5 and reverse the sign:x2− 12x + 20 < 0
    −100−50050100024681012price in dollars, xprofit in dollars, Pa profit: −5x2+ 60x − 100 > 0divide both sides by −5 and reverse the sign:x2− 12x + 20 < 0
    The stall makes a profit when P > 0. Dividing both sides by −5 reverses the sign: x2 − 12x + 20 < 0.

  2. 2.Factorize: two numbers with a product of 20 and a sum of −12 are −2 and −10, so (x − 2)(x − 10) < 0. The profit is exactly zero at the roots x = 2 and x = 10.

    −100−50050100024681012price in dollars, xprofit in dollars, Px = 2x = 10(x − 2)(x − 10) < 0the profit is zero at x = 2 and at x = 10
    −100−50050100024681012price in dollars, xprofit in dollars, Px = 2x = 10(x − 2)(x − 10) < 0the profit is zero at x = 2 and at x = 10
    Factorize: (x − 2)(x − 10) < 0. The profit is exactly zero at x = 2 and at x = 10.

  3. 3.The coefficient of x2 in P is negative, so the graph of P opens downward, and it is above the x-axis between the roots. Test a price in each part: at x = 6, P = −180 + 360 − 100 = 80, and at x = 1 and at x = 11, P = −45.

    −100−50050100024681012price in dollars, xprofit in dollars, Px = 2x = 10P = 80P = −45P = −45the graph opens downward: above the axis between the rootsx = 6 gives P = 80; x = 1 and x = 11 give P = −45
    −100−50050100024681012price in dollars, xprofit in dollars, Px = 2x = 10P = 80P = −45P = −45the graph opens downward: above the axis between the rootsx = 6 gives P = 80; x = 1 and x = 11 give P = −45
    The graph of P opens downward, so it is above the x-axis between the roots: P = 80 at x = 6, and P = −45 at x = 1 and at x = 11.

  4. 4.(a) The stall makes a profit when 2 < x < 10, that is, when a cup costs more than $2 and less than $10.

    −100−50050100024681012price in dollars, xprofit in dollars, Px = 2x = 102 < x < 10a profit when 2 < x < 10more than $2 and less than $10 a cup
    −100−50050100024681012price in dollars, xprofit in dollars, Px = 2x = 102 < x < 10a profit when 2 < x < 10more than $2 and less than $10 a cup
    (a) The stall makes a profit when 2 < x < 10.

  5. 5.(b) The highest point of the graph is halfway between the roots, at x = 2 + 102 = 6, where P = 80. The greatest profit is $80 a day, at a price of $6 a cup.

    −100−50050100024681012price in dollars, xprofit in dollars, Px = 2x = 102 < x < 10(6, 80)the highest point is halfway between the roots: x = 6P = −180 + 360 − 100 = 80: a profit of $80
    −100−50050100024681012price in dollars, xprofit in dollars, Px = 2x = 102 < x < 10(6, 80)the highest point is halfway between the roots: x = 6P = −180 + 360 − 100 = 80: a profit of $80
    (b) The highest point is halfway between the roots, at x = 6, where the profit is $80.

Answer: (a) 2 < x < 10: a price of more than $2 and less than $10; (b) a price of $6, which gives a profit of $80

Common mistakes

  • Dividing by −5 and keeping the sign as >. Dividing both sides of an inequality by a negative number reverses the sign, so x2 − 12x + 20 must be less than zero.
  • Answering x < 2 or x > 10. That is where x2 − 12x + 20 is positive, which is where the profit is negative. A test price such as x = 6, where P = 80, shows which part of the number line is wanted.

More quadratic equations problems, worked step by step →

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