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    Not sure where to start
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    (Original post by poppydoodle)
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    Not sure where to start
    Consider how many points of intersection there will be between the two lines given that the straight line is a tangent to the curve. Then when you find the x coordinate of the intersection, make sure that it has the right number of solutions. That will then tell you what value k needs to take.
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    (Original post by Towcestermaths)
    Consider how many points of intersection there will be between the two lines given that the straight line is a tangent to the curve. Then when you find the x coordinate of the intersection, make sure that it has the right number of solutions. That will then tell you what value k needs to take.
    i know it’s got one point of intersection but idk how to work it out
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    (Original post by poppydoodle)
    i know it’s got one point of intersection but idk how to work it out
    Do you know how to work out the point of intersection?
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    (Original post by Towcestermaths)
    Do you know how to work out the point of intersection?
    y=x^2 - 4x +k - 1
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    (Original post by Carolyne2891)
    make y=y and solve for x, then sub in the x value for y
    i’ve got to y=x^2 - 4x +k - 1
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    (Original post by poppydoodle)

    Not sure where to start
    I'm not sure why others are giving you hints unrelated to differentiation, perhaps they can't read the title properly.

    Look, for the first part, you know that they are tangent and some point. At that point they MUST have the same gradient. You know one of the equations is y=1-3x which has gradient -3 throughout, so just find the point on the parabola with gradient = -3

    Then for part b, you can work out the point where they're tangent using the line. Then just sub the coordinates into the parabola and work out k
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    Here's how I would do it:

    First, you know that at the point where the tangent meets the curve, the curve has the same gradient as the tangent (-3).

    Differentiate the curve. The +k bit at the end disappears, and you should be left as a linear function. Set it equal to -3 and solve for x.

    Once you know the value of x, find the value of y by using the tangent's equation. Then substitute both x and y back in to find k.
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    (Original post by Carolyne2891)
    make y=y and solve for x, then sub in the x value for y
    I don't think that will work because you'll have an equation containing both x and k.
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    Differnetiate and solve it for 3 (As the line as gradient 3).

    Any other way is to set them equal to each other and make b^2 - 4ac = 0 (one pt. of intersection)

    Both methods are acceptable I think.
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    (Original post by poppydoodle)
    i’ve got to y=x^2 - 4x +k - 1
    Thats right (at least your expression = 0 is correct). Now if you use the quadratic formula to calculate the possible values of x you should be able to see what value of k will only give one solution. Hint: think about the discriminant.
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    (Original post by TheMindGarage)
    Here's how I would do it:

    First, you know that at the point where the tangent meets the curve, the curve has the same gradient as the tangent (-3).

    Differentiate the curve. The +k bit at the end disappears, and you should be left as a linear function. Set it equal to -3 and solve for x.

    Once you know the value of x, find the value of y by using the tangent's equation. Then substitute both x and y back in to find k.
    i got f’(x) = 2x+7. how do i set it equal to -3
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    (Original post by Towcestermaths)
    Thats right (at least your expression = 0 is correct). Now if you use the quadratic formula to calculate the possible values of x you should be able to see what value of k will only give one solution. Hint: think about the discriminant.
    Ive just read RDKs reply and he is quite right. I didn't read the title of your question. The method I have suggested will give you the correct answer, but if you need to use differentiation to solve the problem, then his method is the correct one.
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    (Original post by RDKGames)
    I'm not sure why others are giving you hints unrelated to differentiation, perhaps they can't read the title properly.

    Look, for the first part, you know that they are tangent and some point. At that point they MUST have the same gradient. You know one of the equations is y=1-3x which has gradient -3 throughout, so just find the point on the parabola with gradient = -3

    Then for part b, you can work out the point where they're tangent using the line. Then just sub the coordinates into the parabola and work out k
    thank you could you explain how to find the point on the parabola with gradient -3
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    (Original post by poppydoodle)
    thank you could you explain how to find the point on the parabola with gradient -3
    Well it's quote obvious, isn't it? You have (almost) successfully differentiated f(x) to give f'(x)=2x-7. Now you want to find the point with gradient -3, so set f'(x)=-3 to get 2x-7=-3 and solve for x.
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    (Original post by RDKGames)
    Well it's quote obvious, isn't it? You have (almost) successfully differentiated f(x) to give f'(x)=2x-7. Now you want to find the point with gradient -3, so set f'(x)=-3 to get 2x-7=-3 and solve for x.
    2x-4=0?? so x is 2

    sorry my brain doesn’t seem to be working
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    (Original post by poppydoodle)
    2x-4=0?? so x is 2

    sorry my brain doesn’t seem to be working
    Yes that would be the x-coordinate of the point of intersection. Now you want to find the y-coordinate as a start for part b.
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    (Original post by RDKGames)
    Yes that would be the x-coordinate of the point of intersection. Now you want to find the y-coordinate as a start for part b.

    sorry how do you work out y value
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    (Original post by poppydoodle)
    sorry how do you work out y value
    By using the linear equation, perhaps?
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    (Original post by RDKGames)
    By using the linear equation, perhaps?
    -5?
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