Could someone please help me with this question;
Find the tangent to the curve y=x^2-8x+14 which slope is -2. Show that this tangent touches the curve y= -x^2-4x+4. Find the point of contact.
Thank you.
take the derivative of the first equation
y'=2x-8
-2=2x-8
x= 3 is where the tangent has the slope of -2
so the equation of that tangent line is
y=-2x+b to find b, you have to find y at x=3
y= x^2-8x+14=9-24+14
now put that y into
y=-2x + b and solve for b.
Now you have the line. I will leave it up to you to show this line is tangent to the second curve, the process is very similar.
Thank you Bobpursley.
After doing what you told me, the point of contact I got is, (-1,7).
Is this correct?
To find the tangent to the curve y = x^2 - 8x + 14 with a slope of -2, we need to follow these steps:
Step 1: Find the derivative of the curve y = x^2 - 8x + 14.
Step 2: Substitute the slope (-2) into the derivative and solve for x.
Step 3: Substitute the value of x found in Step 2 into the original equation to find the corresponding y-coordinate.
Step 4: Use the equation of the tangent line to determine the equation of the tangent that touches the curve y = x^2 - 8x + 14.
Step 5: Substitute the equation of the tangent into the equation y = -x^2 - 4x + 4 and solve for the x-coordinate to find the point of contact.
Step 6: Substituting the x-value obtained in Step 5 into the equation of the tangent will give the corresponding y-coordinate.
Now let's go through these steps one by one:
Step 1: Find the derivative of the curve y = x^2 - 8x + 14.
The derivative of y = x^2 - 8x + 14 with respect to x is:
dy/dx = 2x - 8.
Step 2: Substitute the slope (-2) into the derivative and solve for x.
Setting the derivative equal to the given slope -2:
2x - 8 = -2.
Solving for x:
2x = -2 + 8,
2x = 6,
x = 6/2 = 3.
So, the x-coordinate of the point where the tangent touches the curve is x = 3.
Step 3: Substitute the value of x found in Step 2 into the original equation to find the corresponding y-coordinate.
Using the original equation y = x^2 - 8x + 14, substitute x = 3:
y = 3^2 - 8(3) + 14,
y = 9 - 24 + 14,
y = -1.
Therefore, the point of contact between the tangent and the curve y = x^2 - 8x + 14 is (3, -1).
Step 4: Use the equation of the tangent line to determine the equation of the tangent that touches the curve y = x^2 - 8x + 14.
Since we know the slope (-2) and a point (3, -1) on the tangent line, we can use the point-slope formula:
y - y1 = m(x - x1), where m is the slope and (x1, y1) is the given point.
Substituting in the values:
y - (-1) = -2(x - 3),
y + 1 = -2x + 6,
y = -2x + 5.
So, the equation of the tangent that touches the curve is y = -2x + 5.
Step 5: Substitute the equation of the tangent into the equation y = -x^2 - 4x + 4 and solve for the x-coordinate to find the point of contact.
Substituting y = -2x + 5 into y = -x^2 - 4x + 4:
-2x + 5 = -x^2 - 4x + 4.
Rearranging the equation:
x^2 - 2x + 4x - 5 - 4 = 0,
x^2 + 2x - 9 = 0.
Factoring the quadratic equation:
(x + 3)(x - 3) = 0.
Setting each factor equal to zero:
x + 3 = 0, x - 3 = 0.
Solving for x:
x = -3 or x = 3.
So, the x-coordinate of the point of contact is either x = -3 or x = 3.
Step 6: Substituting the x-value obtained in Step 5 into the equation of the tangent will give the corresponding y-coordinate.
Using the equation of the tangent, y = -2x + 5, if x = 3:
y = -2(3) + 5,
y = -6 + 5,
y = -1.
Therefore, the point of contact between the tangent and the curve y = -x^2 - 4x + 4 is (3, -1).
To summarize:
The tangent to the curve y = x^2 - 8x + 14 with a slope of -2 touches the curve y = -x^2 - 4x + 4 at the point (3, -1).