The graph of $y = 2x^2 + 3x - 9$ is drawn below - OCR - GCSE Maths - Question 19 - 2020 - Paper 6

We know that the line $y = ax + b$ intersects the curve $y = 2x^2 + 3x - 9$. For the line to intersect the graph at the vertex, we require the slope of the line (represented by $a$) to equal the slope of the tangent to the curve at that point. The vertex of the curve can be found using:

$x_v = -\frac{b}{2a} = -\frac{3}{2 \cdot 2} = -\frac{3}{4}$

The corresponding $y$-coordinate is:

$y_v = 2(-\frac{3}{4})^2 + 3(-\frac{3}{4}) - 9 = -\frac{43}{8}$

To assume the line is in the form $y = -\frac{3}{4}x + b$ and using the vertex point to find $b$ gives us:

$b = y_v + \frac{3}{4}(-\frac{3}{4}) = -\frac{43}{8} + \frac{9}{16} = -\frac{83}{16}$

So, $a = \frac{3}{4}$ and $b = -\frac{83}{16}$.

To solve this new equation using the graph, we again look for the points where the graph of $y = 2x^2 + 3x - 9$ intersects the line $y = ax + b$. We can adjust the line accordingly based on the slope and y-intercept calculated in part (b).
From the graph, identify the new intersection points; these can be approximately at $x = -2$ and $x = 2$. Hence, the solutions for this equation are:

$x = -2$ or $x = 2$.