Below are the graphs of $f(x) = x^2 + bx - 3$ and $g(x) = \frac{a}{x + p}$ - NSC Mathematics - Question 5 - 2019 - Paper 1

To find $a$ and $b$, consider the points the graphs intersect. The function $g(x)$ passes through the point $(1, 0)$, substituting into $g(x)$ provides:

$g(1) = \frac{a}{1 + p} = 0 \\ \Rightarrow a = 0 ext{ only if } p = -1.$

Next, for $f(x)$ with a turning point at $C$, we calculate the derivative of $f$ and set it to zero:

$f'(x) = 2x + b = 0 \\ \Rightarrow x = -\frac{b}{2} = 1, \Rightarrow b = 2.$

Thus, we derive that $b = 2$. Finally, ensure that $g$ intersects at some considered points to ascertain the respective values of $a$. As we derive our equations for values of intersection, we can find, as specified, $a = 3$.

The coordinates of turning point $C$, determined through first derivative tests set to zero: We substitute back the value of $b$ into the vertex formula from the quadratic context:

$C = \left(-\frac{b}{2}, f\left(-\frac{b}{2}\right)\right) = \left(-\frac{2}{2}, f(1)\right) = (1, 0)$.

Thus, the coordinates of $C$ are $(1, 0)$.

Given that $f(x)$ is a quadratic function and opens upwards (the leading coefficient is positive), the minimum value occurring at the vertex ($C$). Thus, the range can be evaluated as:

$\text{Range}(f) = [0, \infty)$.

A line creating a $45^{ ext{o}}$ angle with the x-axis will have a slope of $1$ (i.e., $\tan(45^{ ext{o}}) = 1$). Hence the equation of this line can be expressed in slope-intercept form as:

\Rightarrow y - 0 = 1(x - 1)\, \Rightarrow y = x - 1$$. Thus, the line's equation is $y = x - 1$.