Find a negative value of k so that the graph of y = x2 - 2x + 7 and the graph of y = kx + 5 are tangent?
It says the answer is -2-2sqt2
I graphed this and it didn't look tangent to me. Thanks.
Ok so I got x^(2) (-2-k)x + 2 = 0
Ultimately I got k^(2) +4k -4 = 0 I used quadratic formula and came up with the correct answer. So theoretically, when you subtract the two equations, the solution(s) of that equation is the intersection(s) of both equations? And with resepect to the calculator, did I plug it on wrongly? Why did they not appear to intersect?
I would be careful with terminology. First, one can only talk about the intersection of graphs (which are geometrical figures), not equations. Second, we can draw all points satisfying two equations $\displaystyle f_1(x,y)=0$ and $\displaystyle f_2(x,y)=0$, but this would not be the set of points satisfying $\displaystyle f_1(x,y)-f_2(x,y)=0$. For example, points satisfying 2x + 3y = 0 and x + y = 0 are lines that intersect at just one point (0, 0), but points satisfying x + 2y = 0 fill a whole line. What we have in the original problem are equations $\displaystyle y = x^2 - 2x + 7$ and $\displaystyle y = kx + 5$, to be sure, but not arbitrary equations. Rather, they are definitions of functions with only y in the left-hand side.
So, the correct statement is, If you have functions $\displaystyle y=f_1(x)$ and $\displaystyle y=f_2(x)$, then the solutions of the equation $\displaystyle f_1(x)-f_2(x)=0$ give the (x-coordinates of the) intersection points of the graphs of the two functions.
I don't have a graphing calculator, so I am not sure. There are many possible reasons. E.g., you may need to adjust the window size, write "2*sqt(2)" instead of "2sqt2," etc. Try using 1.41 instead of sqt2.