Quadratic graphs Worked Solutions — GCSE Maths

Fully worked, step-by-step solutions to GCSE Quadratic graphs questions. See exactly how to solve problems on y-intercept, quadratic graph, table of values, substitution.

y-interceptquadratic graphtable of valuessubstitutionparabola shapesign of a
GCSE Foundation70 questionsStep-by-step solutions
Question 1
1 markeasy
The graph of y=x2+3x+4y = x^2 + 3x + 4 is a parabola. Write down the coordinates of the point where the curve crosses the yy-axis.

Worked solution

  1. Use the y-axis

    x=0x = 0

    Every point on the y-axis has x=0x = 0.

  2. Substitute x=0x = 0 into the equation

    y=02+3(0)+4=4y = 0^2 + 3(0) + 4 = 4

    Only the number term is left, so the y-intercept is the value of c.

  3. Write the coordinates

    (0, 4)(0,\ 4)

    The curve crosses the y-axis at (0, 4).

Answer
(0,4)(0, 4)
Question 2
1 markeasy
A table of values is being completed for y=x2−3y = x^2 - 3. Work out the value of yy when x=−2x = -2.

Worked solution

  1. Substitute x=−2x = -2

    y=(−2)2−3y = (-2)^2 - 3

    Replace x with -2, keeping the brackets.

  2. Square the negative number first

    (−2)2=4(-2)^2 = 4

    A negative number squared is positive.

  3. Subtract 3

    y=4−3=1y = 4 - 3 = 1

    So the point (-2, 1) lies on the curve.

Answer
y=1y = 1
Question 3
1 markeasy
Describe the shape of the graph of y=x2+2x−3y = x^2 + 2x - 3.

Worked solution

  1. Look at the coefficient of x squared

    a=1a = 1

    In y=ax2+bx+cy = ax^2 + bx + c the shape is decided by aa.

  2. Decide which way the curve opens

    a=1>0a = 1 > 0

    A positive coefficient of x2x^2 makes the parabola open upwards.

  3. State the shape

    U-shaped parabola\text{U-shaped parabola}

    The curve is a U-shaped parabola with a minimum point.

Answer
AU−shapedparabola,becausethecoefficientofx2ispositive.A U-shaped parabola, because the coefficient of x^2 is positive.
Question 4
2 markseasy
The graph of y=x2−5x+6y = x^2 - 5x + 6 is shown. Use the graph to write down the solutions of x2−5x+6=0x^2 - 5x + 6 = 0.

Worked solution

  1. Solutions are where the curve meets the x-axis

    y=0y = 0

    Solving x2−5x+6=0x^2 - 5x + 6 = 0 means finding the x-values where y=0y = 0.

  2. Read the two crossing points and confirm by factorising

    x2−5x+6=(x−2)(x−3)x^2 - 5x + 6 = (x - 2)(x - 3)

    The curve cuts the x-axis at x=2x = 2 and x=3x = 3, and the factors confirm those roots.

  3. Write down the solutions

    x=2 or x=3x = 2 \text{ or } x = 3

    A parabola that cuts the x-axis twice gives two solutions.

Answer
x=2orx=3x = 2 or x = 3
Question 5
1 markeasy
A point on the curve y=x2+2x−5y = x^2 + 2x - 5 has x=3x = 3. Work out the value of yy.

Worked solution

  1. Substitute x=3x = 3

    y=32+2(3)−5y = 3^2 + 2(3) - 5

    Replace every x with 3.

  2. Work out each term

    32=93^2 = 9

    Powers first, then the multiplication 2×3=62 \times 3 = 6.

  3. Add and subtract

    y=9+6−5=10y = 9 + 6 - 5 = 10

    So the curve passes through (3, 10).

Answer
y=10y = 10

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