Given: $f(x)=x^2-5x-14$ and $g(x)=2x-14$ 5.1 On the same set of axes, sketch the graphs of $f$ and $g$ - NSC Mathematics - Question 5 - 2017 - Paper 1

To find the tangent line, we need the slope of the tangent at x=rac{1}{2} and the point on the curve:

1. Find $f' (x)$:

   $f'(x) = 2x-5$

2. Calculate the slope at x=rac{1}{2}:

   f'(rac{1}{2}) = 2(rac{1}{2}) - 5 = -4

3. Find the value of f(rac{1}{2}):

   f(rac{1}{2}) = (rac{1}{2})^2 - 5(rac{1}{2}) - 14 = -14.25

4. Equation of the tangent line at that point using point-slope form, $y - y_1 = m(x - x_1)$:

   y + 14.25 = -4(x - rac{1}{2})

   Simplifying gives:

   $y = -4x + 2 - 14.25$

   $y = -4x - 12.25$

The quadratic equation $f(x)=k$ can be stated as:

$x^2 - 5x - (14+k) = 0$

For this equation to have two distinct positive roots, the discriminant must be positive:

1. Calculate the discriminant:

   $D = b^2 - 4ac = (-5)^2 - 4(1)(-(14+k)) > 0$

   $25 + 56 + 4k > 0$

   Simplifying:

   $4k > -81$

   Thus,

   k > -rac{81}{4}

2. Ensure roots are positive:

   • The roots of the quadratic given by the quadratic formula are:

   x = rac{-b ext{±} ext{ } ext{sqrt{D}}}{2a}

   • For both roots to be positive, set the vertex criteria:

   -rac{b}{2a} = rac{5}{2} ext{ and } 14 + k < 0

   • This gives us:

   $k < -14$

In summary, the values of k must satisfy:

-rac{81}{4} < k < -14

1. Reflecting $g$ in the x-axis:

   • When reflecting the graph of $g(x)=2x-14$, we get:

   $y = -g(x) = -2x + 14$

2. Translating it 7 units to the left:

   • Replace $x$ with $x + 7$:

   $y = -2(x + 7) + 14$

   • Expanding the equation gives:

   $y = -2x - 14 + 14 = -2x$

Thus, the equation of the graph $h$ is:

$h(x) = -2x$