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Find the gradients of the following lines and state whether the lines are constant, increasing or decreasing functions:

Mathematics

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Step-by-step solution

7 steps
  1. Calculate the gradient for 10.1.

    We use the points (−2,0)(-2, 0) and (2,3)(2, 3).
    The gradient formula is m=y2−y1x2−x1m = \frac{y_2 - y_1}{x_2 - x_1}.

    m=3−02−(−2)=32+2=34m = \frac{3 - 0}{2 - (-2)} = \frac{3}{2 + 2} = \frac{3}{4}

    Since the gradient is positive, the function is increasing.
    The gradient for 10.1 is 34\boxed{\frac{3}{4}}, and it is an increasing function.

  2. Calculate the gradient for 10.2.

    We use the points (−4,3)(-4, 3) and (4,3)(4, 3).

    m=3−34−(−4)=04+4=08=0m = \frac{3 - 3}{4 - (-4)} = \frac{0}{4 + 4} = \frac{0}{8} = 0

    Since the gradient is zero, the function is constant.
    The gradient for 10.2 is 0\boxed{0}, and it is a constant function.

  3. Calculate the gradient for 10.3.

    We use the points (−1,0)(-1, 0) and (0,−1)(0, -1).

    m=−1−00−(−1)=−11=−1m = \frac{-1 - 0}{0 - (-1)} = \frac{-1}{1} = -1

    Since the gradient is negative, the function is decreasing.
    The gradient for 10.3 is −1\boxed{-1}, and it is a decreasing function.

  4. Determine the equation for line g in 11.1.

    We use the points (0,3)(0, 3) and (2,0)(2, 0).
    First, calculate the gradient:

    m=0−32−0=−32m = \frac{0 - 3}{2 - 0} = \frac{-3}{2}

    The y-intercept is c=3c = 3 (from the point (0,3)(0, 3)).
    Using the slope-intercept form y=mx+cy = mx + c:

    y=−32x+3y = -\frac{3}{2}x + 3

    The equation for line g is y=−32x+3\boxed{y = -\frac{3}{2}x + 3}.

  5. Determine the equation for line h in 11.1.

    Line h is a horizontal line passing through y=−1y = -1.
    The equation for line h is y=−1\boxed{y = -1}.

  6. Determine the equation for 11.2.

    We use the points (−3,0)(-3, 0) and (0,−4)(0, -4).
    First, calculate the gradient:

    m=−4−00−(−3)=−43m = \frac{-4 - 0}{0 - (-3)} = \frac{-4}{3}

    The y-intercept is c=−4c = -4 (from the point (0,−4)(0, -4)).
    Using the slope-intercept form y=mx+cy = mx + c:

    y=−43x−4y = -\frac{4}{3}x - 4

    The equation for 11.2 is y=−43x−4\boxed{y = -\frac{4}{3}x - 4}.

  7. Determine the equation for 11.3.

    We use the points (−4,−2)(-4, -2) and (0,2)(0, 2).
    First, calculate the gradient:

    m=2−(−2)0−(−4)=2+20+4=44=1m = \frac{2 - (-2)}{0 - (-4)} = \frac{2 + 2}{0 + 4} = \frac{4}{4} = 1

    The y-intercept is c=2c = 2 (from the point (0,2)(0, 2)).
    Using the slope-intercept form y=mx+cy = mx + c:

    y=1x+2y = 1x + 2

    The equation for 11.3 is y=x+2\boxed{y = x + 2}.

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