1. geometrical interpretations of the solutuon of equations- HELP PLEASE

Hey this is probably really simple but I can't get my head around it.

1) Find the possible values of a for which y = ax+1 is a tangent to y = 3-4x+4

2) Find the possible values of k for which y = x-8 is a tangent to y= 4x²+ kx + 1

Would be really grateful if someone could just get me started and explain what to do. Thanks

2. Originally Posted by lra11
Hey this is probably really simple but I can't get my head around it.

1) Find the possible values of a for which y = ax+1 is a tangent to y = 3-4x+4
I'll get you started with the first. the second should be similar.

what do we know about the line: $y = ax + 1$?

it has slope $a$ and y-intercept 1

how do we find the slope? by the derivative, so let's find the derivative. i will call the slope m.

$m = 6x - 4$ ..................(1)

ok, now we know that the given line passes through (0,1) and that all points on the curve are given by $\left( x, 3x^2 - 4x + 4 \right)$

thus, the tangent line must pass through the points $( \underbrace{0}_{x_1} , \underbrace{1}_{y_1})$ and $( \underbrace{x}_{x_2} , \underbrace{3x^2 - 4x + 4}_{y_2} )$

so another formula from the slope is (from precalc):

$m = \frac {y_2 - y_1}{x_2 - x_1} = \frac {3x^2 - 4x + 3}{x}$ ....................(2)

but we are talking about the same lines here, the slopes must be equal. thus we equate (1) and (2) to obtain:

$6x - 4 = \frac {3x^2 - 4x + 3}x$

solving this, we find that $x = \pm 1$

thus, the points any tangent line would go through are:

$(0,1)$ and $(1,3)$ ....tangent line 1

or

$(0,1)$ and $(-1,11)$ ....tangent line 2

find the two lines that pass through those points and you will find the two values for $a$

3. Originally Posted by lra11
Hey this is probably really simple but I can't get my head around it.

1) Find the possible values of a for which y = ax+1 is a tangent to y = 3-4x+4
For a parabola "being tangent" is the same as saying "only one intersection point".

Thus,
$ax+1 = 3x^2 - 4x+4$ has only a single solution.
Thus,
$3x^2 - (4+a)x+3 = 0$
To have exactly one solution we require that,
$(4+a)^2 - 4(3)(3) = 0$

4. Originally Posted by ThePerfectHacker
For a parabola "being tangent" is the same as saying "only one intersection point".

Thus,
$ax+1 = 3x^2 - 4x+4$ has only a single solution.
Thus,
$3x^2 - (4+a)x+3 = 0$
To have exactly one solution we require that,
$(4+a)^2 - 4(3)(3) = 0$
well, that was easy. i need a Staples easy button

5. thanks

thanks for the help both of you