Sketch the graph of y = 2x² - 8x - 5 showing the coordinates of the turning point and the exact coordinates of any intercepts with the coordinate axes. - Edexcel - GCSE Maths - Question 21 - 2019 - Paper 1

Set y = 0 to find the x-intercepts:

$0 = 2x^2 - 8x - 5$

Using the quadratic formula, where a = 2, b = -8, c = -5:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

Calculating the discriminant:

$b^2 - 4ac = (-8)^2 - 4(2)(-5) = 64 + 40 = 104$

Thus, we have:

$x = \frac{8 \pm \sqrt{104}}{4} = 2 \pm \frac{\sqrt{26}}{2}$

This gives us the x-intercepts at the points $(2 + \frac{\sqrt{26}}{2}, 0)$ and $(2 - \frac{\sqrt{26}}{2}, 0)$.

The turning point can be found using the vertex formula x = -\frac{b}{2a}:

$x = \frac{8}{2(2)} = 2$

To find the y-coordinate of the turning point, substitute x back into the equation:

$y = 2(2)^2 - 8(2) - 5 = 8 - 16 - 5 = -13$

Thus, the turning point is at (2, -13).

In the sketch:

• Plot the y-intercept at (0, -5).
• Plot the x-intercepts at $(2 + \frac{\sqrt{26}}{2}, 0)$ and $(2 - \frac{\sqrt{26}}{2}, 0)$.
• Plot the turning point at (2, -13).
• Ensure the parabola opens upwards and the features are clearly labeled.