Given that $f(x) = 2x^2 + 8x + 3$ (a) find the value of the discriminant of $f(x).$ (b) Express $f(x)$ in the form $p(x + q)^2 + r$ where $p, q$ and $r$ are integers to be found - Edexcel - A-Level Maths Pure - Question 1 - 2014 - Paper 1

We start with the quadratic function:

$f(x) = 2x^2 + 8x + 3$

We can factor out the 2 from the first two terms:

$f(x) = 2(x^2 + 4x) + 3$

Next, we complete the square for the expression inside the parentheses:

1. Take half of the coefficient of $x$ (which is 4), square it to get $(2)^2 = 4$.
2. Add and subtract this square inside the parentheses:

$f(x) = 2(x^2 + 4x + 4 - 4) + 3$ $f(x) = 2((x + 2)^2 - 4) + 3$ $f(x) = 2(x + 2)^2 - 8 + 3$ $f(x) = 2(x + 2)^2 - 5$

Thus, the expression for $f(x)$ in the required form is: $f(x) = 2(x + 2)^2 - 5$ Here, $p = 2$, $q = 2$, and $r = -5$.

Given the line $y = 4x + c$ is a tangent to the curve $y = f(x) = 2x^2 + 8x + 3$, we need to find the point of tangency.

First, we differentiate $f(x)$ to find the slope at any point on the curve:

f'(x) = rac{d}{dx}(2x^2 + 8x + 3) = 4x + 8

For the line to be tangent to the curve, the slope of the curve at the point of tangency must equal the slope of the line, which is 4:

Setting the derivative equal to 4:

$4x + 8 = 4$ $4x = -4$ $x = -1$

Now, we substitute $x = -1$ into the original function to find the corresponding $y$ value:

$f(-1) = 2(-1)^2 + 8(-1) + 3 = 2 - 8 + 3 = -3$

At this point of tangency, the coordinates are $(-1, -3)$.

Substituting $x = -1$ into the line equation to find $c$:

$y = 4(-1) + c = -3$ $-4 + c = -3$ So, $c = -3 + 4 = 1$

Thus, the value of $c$ is $1$.