# Thread: tangent

1. ## tangent

Given that the curve y = 5 + 4x -x^2 has a tangent equation in the form of y = mx + 9. Calculate the values of m without drawing a graph or using differentiation.

2. Originally Posted by z1llch
Given that the curve y = 5 + 4x -x^2 has a tangent equation in the form of y = mx + 9. Calculate the values of m without drawing a graph or using differentiation.
the tangent line must have a point in common with the graph. so since $\displaystyle y = 5 + 4x - x^2$ on the graph, the $\displaystyle y$ in $\displaystyle y = mx + 9$ must have the same value, so

$\displaystyle 5 + 4x - x^2 = mx + 9$

moreover, the tangent line must touch only ONE point on the graph. so the $\displaystyle m$ we want occurs when the discriminant of the above quadratic is zero

3. Hello, z1llch!

Given that the curve $\displaystyle y \:= \:5 + 4x -x^2$ has a tangent equation in the form: $\displaystyle y \:= \:mx + 9$,
calculate the values of $\displaystyle m$ without drawing a graph or using differentiation.

If that line tangent to the parabola, they intersect at one point.

We have: .$\displaystyle mx + 9 \:=\:5 + 4x - x^2 \quad\Rightarrow\quad x^2 + (m-4)x + 4 \:=\:0$

Quadratic Formula: .$\displaystyle x \;=\;\frac{-(m-4) \pm\sqrt{m^2-8m}}{2}$

There will be one point of intersection if the disriminant is zero.

Hence: .$\displaystyle m^2-8m \:=\:0\quad\Rightarrow\quad m(m-8) \:=\:0\quad\Rightarrow\quad\boxed{ m \:=\:0,\:8}$