Help with graph questions?

Question

The Q says the curve C has equation y = 4x^2 - 7x +11 and the line L has equation y = 5x + k
Where k is a constant. Given that L intersects C in two distinct points show that k > 2

Ive tried everything and cannot get to the answer. please help!

Answer 1

To show this, set the 2 equations equal, bring everything to one side, and compute the value of k that makes the discriminant greater than 0. That will show that the graphs meet in 2 points. (When the discriminant is 0, they meet in 1 point; when less than 0, no points.)

5x + k = 4x² - 7x + 11
0 = 4x² - 12x + (11 - k)

This is a quadratic equation, with a = 4, b = -12, and c = (11 - k).

You want the discriminant, b² - 4ac, to be greater than 0.

b² - 4ac > 0
(-12)² - 4(4)(11 - k) > 0
144 - 176 + 16k > 0
16k > 32
k > 2

Answer 2

Try setting the two equation equivalent to each other. You'll end up with something you can solve with the quadratic formula.

In the case of one root to the quadratic formula you have 1 intersection. With two real roots you have 2 intersections. And with two imaginary roots, you have no intersections...

C: y = 4x² - 7x + 11
L: y = 5x + k

Now set them equivalent:
4x² - 7x + 11 = 5x + k

Now subtract 5x + k to get everything on the left:
4x² - 7x + 11 - 5x - k = 0

And simplify:
4x² - 12x + (11 - k) = 0

Divide both sides by 4:
x² - 3x + (11 - k)/4 = 0

Now you can complete the square by taking the x-coefficient (-3), halving it (-3/2) and squaring it (9/4). Add this and subtract it:
x² - 3x + 9/4 - 9/4 + (11 - k)/4 = 0

Turn the first part into a perfect square:
(x - 3/2)² + [-9/4 + (11 - k)/4] = 0

Finally, simplify the item in the brackets:
(x - 3/2)² + (2-k)/4 = 0

Because (x - 3/2)² is always greater than or equal to zero, the single solution will be when (x - 3/2)² is zero and the added expression (2-k)/4 is also zero. This happens when x = 3/2 and k = 2.

However, you want two solutions. Because the first part can't be negative, the only choice is for it to be positive. Then the expression (2-k)/4 would have to be negative to make zero.

Thus you've simplified this all to:
You will have two intersections when (2-k)/4 < 0

And if you solve it further:
2-k < 0
2 < k
k > 2

Q.E.D.