Given that a and b are positive constants, (a) on separate diagrams, sketch the graph with equation (i) y = |2x - a| (ii) y = |2x - a| + b Show, on each sketch, the coordinates of each point at which the graph crosses or meets the axes - Edexcel - A-Level Maths Pure - Question 7 - 2017 - Paper 4

For the equation $y = |2x - a| + b$, the graph is similar to part (i) but shifted upwards by $b$.

• y-axis: Set $x = 0$: $y = |2(0) - a| + b = a + b \Rightarrow (0, a + b)$

• x-axis: Set $y = 0$ to find the x-intercepts: $0 = |2x - a| + b \Rightarrow |2x - a| = -b$ Since $b$ is a positive constant, this equation has no solution, indicating that the graph does not intersect the x-axis. However, we can find the vertex point: $\left(\frac{a}{2}, b\right)$.

Thus the graph meets the y-axis at $(0, a + b)$ and does not cross the x-axis.

Given the equation:

$|2x - a| + b = \frac{3}{2} x + 8$

Substituting $x = 0$: $|2(0) - a| + b = \frac{3}{2}(0) + 8$ | - a | + b = 8 \\ ext{(therefore, $a + b = 8$ if $a$ is positive)}

To find $c$, substitute $x = c$: $|2c - a| + b = \frac{3}{2}c + 8$

Solving for $|2c - a|$: $|2c - a| = \frac{3}{2}c + 8 - b$

Since $b = 8 - a$: $|2c - a| = \frac{3}{2}c + 8 - (8 - a)$ $|2c - a| = \frac{3}{2}c + a$

This gives two cases to solve for $c$:

1. Case 1: $2c - a = \frac{3}{2}c + a$

   $2c - \frac{3}{2}c = 2a \Rightarrow \frac{1}{2}c = 2a \Rightarrow c = 4a$

2. Case 2: $2c - a = -\left(\frac{3}{2}c + a\right)$

   $2c - a = -\frac{3}{2}c - a \\ 2c + \frac{3}{2}c = 0 \Rightarrow \frac{7}{2}c = 0 \Rightarrow c = 0$

Thus, $c = 4a$.