consider the line y=6x-k and the parabola y=x^2
i) for what value of k is the line y=6x-k a tangent to the parabola y=x^2 ?
ii) the line y=6x-k intersects the parabola in two distinct places. what is the largest integer value that k can take ?
i. at x=3, the slope of the parabola is 6, so at 3, the parabola has a y value of 9, which means if the line y=6x+k is =9, then k=....
i)
using Calculus, this is easy
if y = x^2
dy/dx = 2x
but the slope of the line is 6
so 2x = 6
x = 3
if x=3 then y=9 from the parabola
so the point (3,9) must also be on the line
9 = 6(3)-k
k = 9
ii) let's intersect
y = x^2 and y = 6x-k
then x^2 - 6x + k = 0
to have 2 distinct roots, the discriminat must be positive
so b^2 - 4ac > 0
36 - 4(k) > 0
k < 9
since k=9 produces the tangent, and you asked for the largest integer that k could have,
k = 8
(strange question!)
To answer these questions, we need to find the points of tangency and intersection between the line and the parabola. We'll break it down step by step:
i) To find the value of k for which the line y = 6x - k is a tangent to the parabola y = x^2, we need to find the points where the line and the parabola have equal slopes.
The slope of the line y = 6x - k is given by its coefficient of x, which is 6.
The slope of the parabola y = x^2 can be found by taking the derivative of y with respect to x. Differentiating y = x^2, we get dy/dx = 2x.
Setting the slopes equal to each other, we have:
6 = 2x
Solving for x, we get x = 3. Substituting this value of x back into either the line or the parabola equation gives us the corresponding y-coordinate.
If we substitute x = 3 into the line equation y = 6x - k, we have:
y = 6(3) - k
y = 18 - k
This is the y-coordinate of the point of tangency. Since the line is tangent to the parabola, the x-coordinate will also be the same. Therefore, the point of tangency is (3, 18 - k).
Now we substitute x = 3 into the parabola equation y = x^2:
y = (3)^2
y = 9
Therefore, the point of tangency is (3, 9).
Since the line y = 6x - k is a tangent to the parabola, the points of tangency are the same. So, equating the y-coordinates:
18 - k = 9
Solving for k, we get k = 9.
Thus, for the line y = 6x - k to be a tangent to the parabola y = x^2, k must be equal to 9.
ii) To find the largest integer value of k for which the line y = 6x - k intersects the parabola y = x^2 in two distinct places, we need to find the discriminant of the quadratic equation formed by setting the line and parabola equal to each other.
Setting the line y = 6x - k equal to the parabola y = x^2, we have:
x^2 = 6x - k
Rearranging, we get:
x^2 - 6x + k = 0
This equation represents a quadratic equation, and it will have two distinct (real) solutions if the discriminant (b^2 - 4ac) is greater than zero.
The discriminant is given by (-6)^2 - 4(1)(k) = 36 - 4k.
To have two distinct solutions, this discriminant should be greater than zero:
36 - 4k > 0
Simplifying, we get:
4k < 36
Dividing both sides by 4, we have:
k < 9
The largest integer value of k that satisfies this inequality is 8.
Hence, the largest integer value that k can take for the line y = 6x - k to intersect the parabola y = x^2 in two distinct places is 8.