y=x^2 and Y=x^2 + 6x -5?

Question

NOW FIND THE EQUATIONS OF THE TWO LINE TANGENT TO THE TWO GRAPHS.
give me the answer and explain how you came to the answer, you do not need to show work, I usually can figure the operation. I have the answer I just want to make sure I getting a response from someone who actually got the right answer

SETting the derivatives equal does not work for some reason.

Answer 1

I don't think that any of the first 3 responders has given you the right answer.
The first didn't try to solve it.
The second was confused about normal lines vs. tangent lines, and also integrated things that should not be integrated.
The third thought you wanted a line tangent to each graph, rather than a line that is tangent to both graphs.

I, too, had to change the problem that you stated in order to make sense of it. I ignored the word "two" since there is only one line that is tangent to the two graphs. (If you have plotted them, you can see that this is the case.)

Let's say that we want to find a line tangent to the first equation, and we'll use x' and y' for its equation (to distinguish from x and y, which refer to points on the curve(s)). Using the slope/intercept form, we'll express this equation as y' = mx' + b.

In order to be tangent to the first equation, it has to pass through a point (x1,y1) that lies on the curve of the first equation, and has to have a slope of 2x1 at that point. So this tangent is y' = 2x1x' - x1^2.

The corresponding equation for a tangent to the second curve through point (x2,y2) is y' = (2x2+6)x' - 5 - x2^2.

In order for these two lines to be one (i.e., to represent a line tangent to both curves), the slopes and the intercepts of the two lines have to be the same. In other words (or in equation form), 2x1 = 2x2+6 and x1^2 = x2^2 + 5. From the first of these equations, x1 = x2 + 3. Substituting this into the second equation, 6x2 + 9 = 5, so x2 = -2/3 and x1 = 7/3.

The tangent line is then y' = 2x1x' - x1^2 = 14x'/3 - 49/9.
The tangent line is also y' = (2x2+6)x' - 5 - x2^2 (based on (x2,y2) and the second curve), which again produces y' = 14x'/3 - 49/9.

Incidentally, setting the derivatives equal, as you said you did, would simply allow you to find the value of x for which both curves have the same slope. But these particular curves don't have the same slope for any one value of x.

Answer 2

i guess you mean the two tangent lines where the graphs intersect!

x^2=x^2 + 6x -5
so 6x-5=0
x=5/6
and y=(5/6)^2=25/36

now
y'= 2x, so the slope of the tangent line at x= 5/6 is 2(5/6)=5/3
Y'=2x+6, so the corresponding slope is: 2(5/6)+6=23/3

so the equations to the tangent lines are:
y-25/36=5/3(x-5/6),
y-25/36=23/3(x-5/6),
respectively

Answer 3

for the first graph
y = x²
we get the slope of this equation
y' = 2x
the slope of the tangent = -1 / 2x
to get the equation of the tangent we solve
∫-1/2x = -½ ln x
y (tangent1) = -½ ln x


for the equation
y = x² + 6x - 5
y'= 2x + 6
the slope of the tangent = -1/(2x + 6 )
to get the tangent equation we solve
∫-1/(2x + 6 ) =
-½ ln (2x + 6 )

Answer 4

its the red dots on the apple jacks!