Question

show that if the line y=kx-k touches the parabola with equation y=x^2 +5x + 10 then k=-1 or k=15

Answer 1

dang that took a while

Answer 2

got it though

Answer 3

ok lets use the point on the graph of
\[y=x^2+5x+10\] at
\[(a,a^2+5a+10)\] and the slope of the tangent line is given by the derivative, namely
\[y'=2x+5\] so at the point
\[(a,a^2+5a+10)\] the slope is
\[2a+5\] now use the point slope formula to get the equation for the line

Answer 4

point is
\[(a,a^2+5a+10)\] slope is
\[2a+5\] equation is
\[y-(a^2+5a+10)=(2a+5)(x-a)\] or
\[y=(2a+5)(x-a)+a^2+5a+10\]

Answer 5

thanks but im seriously confused....

Answer 6

multiply out to get
\[y=(2a+5)x-2a^2-5a+a^2+5a+10\] or
\[y=(2a+5)x-a^2+10\] or even
\[y=(2a+5)x-(a^2-10)\] and therefore
\[k=2a+5=a^2-10\]

Answer 7

then solve the quadratic
\[2a+5=a^2-10\]
\[a^2-2a-15=0\]
\[(a-5)(a+3)=0\]
\[x=5,a=-3\]

Answer 8

if
\[a=5\] then
\[k=2\times 5+5=15\] and if
\[a=-3\] then
\[k=2\times -3+5=-1\]

Answer 9

now that i have written it, ask away. i was confused at first as well

Answer 10

first of all this problem is huge!!!!! and second exactly how did yo get the point to start off on?

Answer 11

i mean the coordinates

Answer 12

i didn't get a specific point, i just called the first coordinate "a"
if the first coordinate is "a" then since we know the function, the second coordinate must be
\[a^2+5a+10\]

Answer 13

any point on the curve
\[y=x^2+5x+10\] must look like
\[(a,a^2+5a+10)\]

Answer 14

oh ok thanks