If a line is tangent to a curve, then its slope is the derivative of the curve at that point.
The equation of the curve is y = x^2 + k. From this equation, we can say that the derivative of this equation dy/dx is 2x. We can equate 2x with the slope of the line y = 6x - 7, which is 6.
2x = 6
x = 3
Then, substitute x = 3 to y = 6x - 7 in order to get y.
y = 6(3) - 7
y = 11
Therefore, the line y = 6x - 7 is tangent to y = x^2 + k at (x, y) = (3, 11). Substitute (x, y) to the curve equation y = x^2 + k.
11 = 3^2 + k
k = 2 (Answer)
You can check by drawing y = x^2 + 2 and y = 6x - 7 on the same coordinate plane.
The equation of the curve is y = x^2 + k. From this equation, we can say that the derivative of this equation dy/dx is 2x. We can equate 2x with the slope of the line y = 6x - 7, which is 6.
2x = 6
x = 3
Then, substitute x = 3 to y = 6x - 7 in order to get y.
y = 6(3) - 7
y = 11
Therefore, the line y = 6x - 7 is tangent to y = x^2 + k at (x, y) = (3, 11). Substitute (x, y) to the curve equation y = x^2 + k.
11 = 3^2 + k
k = 2 (Answer)
You can check by drawing y = x^2 + 2 and y = 6x - 7 on the same coordinate plane.
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Derivative of y = x^2 + k is y = 2x. So at any given x, the curve has slope 2x.
The line y = 6x - 7 has slope 6.
1. Where is 2x = 6? The line has to be tangent there since the slope is 2x and the slope is 6.
2. What is y = 6x - 7 at that value of x? y = x^2 + k is the same value since the tangent line goes through the curve.
The line y = 6x - 7 has slope 6.
1. Where is 2x = 6? The line has to be tangent there since the slope is 2x and the slope is 6.
2. What is y = 6x - 7 at that value of x? y = x^2 + k is the same value since the tangent line goes through the curve.
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The equation
6*x - 7 = x^2 + k
has only one solution if the line is a tangent to the curve.
x^2 - 6*x + (k + 7) = 0
This must be of the form (x + a)*(x + a) = 0 for only one solution
x^2 + 2*a*x + a^2 = 0
Comparing coefficient of x, a = -3: a^2 = 9 = k + 7
Hence k = 2 <<<
6*x - 7 = x^2 + k
has only one solution if the line is a tangent to the curve.
x^2 - 6*x + (k + 7) = 0
This must be of the form (x + a)*(x + a) = 0 for only one solution
x^2 + 2*a*x + a^2 = 0
Comparing coefficient of x, a = -3: a^2 = 9 = k + 7
Hence k = 2 <<<
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y = x^2 + k
dy/dx = 2x = 6 --> x = 3
So the eqn of tangent is:
y - (k + 9) = 6(x - 3)
y = 6x - 18 + (k + 9)
k - 9 = -7
k = 2
dy/dx = 2x = 6 --> x = 3
So the eqn of tangent is:
y - (k + 9) = 6(x - 3)
y = 6x - 18 + (k + 9)
k - 9 = -7
k = 2
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6 times 6 minus seven= k
simplez
simplez
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well, 6x - 7=x^2+k
(x+7)(x-1)
k= -7,1
(x+7)(x-1)
k= -7,1