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Math Help - Finding value to make two graphs tangent

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    Finding value to make two graphs tangent

    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.
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    Re: Finding value to make two graphs tangent

    It does look tangent.

    Hint: consider the difference of the two functions and find when it has a single root.
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    Re: Finding value to make two graphs tangent

    I am putting it in the calculator wrongly. When I evalued K and then put it in the calculator, it looked tangent.

    This is how I am putting it in: "(-2-2sqt2)x+5"
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    Re: Finding value to make two graphs tangent

    Quote Originally Posted by benny92000 View Post
    I am putting it in the calculator wrongly. When I evalued K and then put it in the calculator, it looked tangent.

    This is how I am putting it in: "(-2-2sqt2)x+5"
    You need the line and the parabola to intersect at a single point. Therefore you need the discriminant of x^2 - 2x + 7 - (kx + 5) = 0 to equal zero ....
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    Re: Finding value to make two graphs tangent

    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?
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    Re: Finding value to make two graphs tangent

    Quote Originally Posted by benny92000 View Post
    So theoretically, when you subtract the two equations, the solution(s) of that equation is the intersection(s) of both equations?
    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 f_1(x,y)=0 and f_2(x,y)=0, but this would not be the set of points satisfying 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 y = x^2 - 2x + 7 and 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 y=f_1(x) and y=f_2(x), then the solutions of the equation f_1(x)-f_2(x)=0 give the (x-coordinates of the) intersection points of the graphs of the two functions.

    Quote Originally Posted by benny92000 View Post
    And with resepect to the calculator, did I plug it on wrongly? Why did they not appear to intersect?
    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.
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