Equations, Inequalities & Graphs (Cambridge International IGCSE Further Maths): Exam Questions

Solve the equation $5 \vert 5 x - 2 \vert - 1 = 14$.

The diagram shows the graph of $y=f(x),\mathrm{where}f(x)=2(x+1)^{2}(x--1)$ . Use the graph to solve the inequality $\mathrm{f} ( x ) less-than \text{or} \text{slanted} \text{equal} \text{to} - 1$.

On the axes, sketch the graph of $y = 5 ( x + 1 ) ( 3 x - 2 ) ( x - 2 )$, stating the intercepts with the coordinate axes.

Extract:

On the axes, sketch the graph of y = 5(x+1)(3x-2)(x-2), stating the intercepts with the coordinate axes.

From the mark scheme: The graph is cubic. Solve 5(x+1)(3x-2)(x-2) = 0 to find where the curve will cross the x axis. x = -1, x = 2/3, x = 2 Substitute x = 0 into y = 5(x+1)(3x-2)(x-2) to find where the curve will cross the y axis. y = 5(1)(-2)(-2) y = 20

Hence find the values of $x$ for which $5 ( x + 1 ) ( 3 x - 2 ) ( x - 2 ) > 0$.

On the axes below sketch the graph of $y=-3(x-2)(x-4)(x+1)$, showing the coordinates of the points where the curve intersects the coordinate axes.

Extract:

On the axes below sketch the graph of y = -3(x-2)(x-4)(x+1), showing the coordinates of the points where the curve intersects the coordinate axes.

The graph will be a negative cubic since -3 × x × x × x = -3x³.

The graph crosses the x-axis when y = 0: 0 = -3(x-2)(x-4)(x+1) Either x-2 = 0, x-4 = 0 or x+1 = 0 So x = 2, x = 4 or x = -1

The graph crosses the y-axis when x = 0: y = -3(0-2)(0-4)(0+1) y = -24

The intercepts are: (2, 0), (4, 0), (-1, 0) and (0, -24)

Hence find the values of $x$ for which $-3(x-2)(x-4)(x+1)>0$.

On the axes below, sketch the graph of $y = \left| 5 x - 7 \right|$, showing the coordinates of the points where the graph meets the coordinate axes.

Extract:

On the axes below, sketch the graph of y = |5x - 7|, showing the coordinates of the points where the graph meets the coordinate axes.

The line y = 5x - 7 would have y-intercept (0, -7) and would intersect the x-axis at (7/5, 0)

We need to sketch y = |5x - 7| which means that all the y values are made positive. This means the part of the graph that goes below the x-axis gets reflected in the x-axis.

The final graph is a correct V shape with vertex on positive x-axis at coordinate (7/5, 0) and y-intercept at coordinate (0, 7).

Solve $5 \left| 5 x - 7 \right| - 1 = 14$.

On the axes below, sketch the graph of $y=-(x+2)(x-1)(x-6)$, showing the coordinates of the points where the graph meets the coordinate axes.

Extract:

On the axes below, sketch the graph of y = -(x+2)(x-1)(x-6), showing the coordinates of the points where the graph meets the coordinate axes.

Find where the graph intersects the x axis by substituting in y = 0 0 = -(x+2)(x-1)(x-6) x = -2 x = 1 x = 6

Find the y intercept by substituting in x = 0 y = -(2)(-1)(-6) y = -12

The shape of the graph will be an inverted cubic.

Hence solve $-(x+2)(x-1)(x-6)\leq0$.

Solve the following equation.

$2 \vert x - 1 \vert \text{less} \text{or} \text{equal} \text{than} \vert 2 - x \vert$

Solve the inequality $\left| 4 x - 1 \right| > 9$.

On the axes below, sketch the graph of $y = \left| ( x - 2 ) ( x + 1 ) ( x + 2 ) \right|$showing the coordinates of the points where the curve meets the axes.

The diagram shows the graph of $y=f(x)$, where $f(x)$ is a cubic polynomial. Find $f(x)$.

Extract:

The diagram shows the graph of y = f(x), where f(x) is a cubic polynomial.

Write down the values of $x$ such that $f(x)<0$.

On the axes below, sketch the graph of $y=(x-2)(x+1)(3-x)$, stating the intercepts on the coordinate axes.

Extract:

On the axes below, sketch the graph of y = (x-2)(x+1)(3-x), stating the intercepts on the coordinate axes.

If we expand the brackets of the equation, the highest power of x we would get would be x cubed. Therefore, the graph is cubic. Since one of the x in the brackets is negative, we will have an inverted cubic shape.

Solve (x-2)(x+1)(3-x) = 0 to find where the curve will cross the x axis. x = 2, x = -1, x = 3

Substitute x = 0 into y = (x-2)(x+1)(3-x) to find where the curve will cross the y axis. y = (-2)(1)(3) y = -6

Hence write down the values of $x$ such that $(x-2)(x+1)(3-x)>0$.

Solve $\left| 3 x - 2 \right| = 4 + x$ .

The graph $y = \mathrm{p} ( x )$ where $\mathrm{p} ( x ) = 5 x^{3} - 17 x^{2} + 16 x + a$ intersects the $x$-axis at the points $( b , 0 )$, $( 1 , 0 )$ and $( c , 0 )$ where $b<c$.

Find the constants $a$, $b$ and $c$.

The curve $y = \mathrm{p} ( x )$ has a local maximum point at $( \frac{2}{3} , \frac{16}{27} )$ and a local minimum point at $( \frac{8}{5} , - \frac{36}{25} )$.

Find the value of $d$ for which the equation $\vert p ( x ) \vert = d$ has exactly 5 distinct solutions.

The diagram shows the graph of $y = \vert \mathrm{f} ( x ) \vert$, where $f(x)$ is a cubic function. Find the possible expressions for $f(x)$ in factorised form.

The diagram shows the graph of a cubic curve .

Find an expression for $\mathrm{f} ( x )$.

Extract:

The diagram shows the graph of a cubic curve.

The graph crosses the x axis at the roots of the equation: x = -2, x = -1, x = 5

f(x) = -1/2(x + 2)(x + 1)(x - 5)

Solve $f(x)\leq0$.

Write $2x^{2}+3x-4$ in the form $a(x+b)^{2}+c$, where $a,b$ and $c$ are constants.

Extract:

Write 2x² + 3x - 4 in the form a(x + b)² + c, where a, b and c are constants.

Answer: 2(x + 3/4)² - 41/8

Hence write down the coordinates of the stationary point on the curve $y=2x^{2}+3x-4$.

Extract:

2(x + 3/4)² - 41/8

On the axes below, sketch the graph of $y = \left| 2 x^{2} + 3 x - 4 \right|$, showing the exact values of the intercepts of the curve with the coordinate axes.

Extract:

The minimum point on the curve y = 2x² + 3x - 4 is (-3/4, -41/8).

The graph of y = |2x² + 3x - 4| is a parabola where everything below the x-axis is reflected above the x-axis. The x-intercepts are ((-3 + √41)/4, 0) and ((-3 - √41)/4, 0), and the y-intercept is (0, 4).

Find the value of $k$ for which $\left| 2 x^{2} + 3 x - 4 \right| = k$ has exactly 3 values of $x$.

The diagram shows the graph of $y = \left| \mathrm{p} ( x ) \right|$ , where $p(x)$ is a cubic function. Find the two possible expressions for $p(x)$.

Solve the inequality $\left| 3 x + 2 \right| > 8 + x$.