tangent

**z1llch** · October 1st 2008, 08:46 AM

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.

**Jhevon** · October 1st 2008, 08:53 AM

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 $y = 5 + 4x - x^2$ on the graph, the $y$ in $y = mx + 9$ must have the same value, so

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

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

**Soroban** · October 1st 2008, 09:01 AM

Hello, z1llch!

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

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

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

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

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

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

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