Sketch the graph of y = 2x^2 - 8x - 5 showing the coordinates of the turning point and the exact coordinates of any intercepts with the coordinate axes. - Edexcel - GCSE Maths - Question 22 - 2019 - Paper 1

To find the y-intercept, set x = 0:

$y = 2(0)^2 - 8(0) - 5 = -5$

Thus, the y-intercept is at the point (0, -5).

For the x-intercepts, set y = 0:

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

Using the quadratic formula, where $a = 2$, $b = -8$, and $c = -5$, the formula is given by:

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

Calculating:

$x = \frac{8 \pm \sqrt{(-8)^2 - 4(2)(-5)}}{2(2)} = \frac{8 \pm \sqrt{64 + 40}}{4} = \frac{8 \pm \sqrt{104}}{4} = \frac{8 \pm 2\sqrt{26}}{4} = 2 \pm \frac{\sqrt{26}}{2}$

Thus, the x-intercepts are at the points ( \left( 2 - \frac{\sqrt{26}}{2}, 0 \right) ) and ( \left( 2 + \frac{\sqrt{26}}{2}, 0 \right) ).