calculus problem: tangent lines?

Question

If the line y = 4x + 3 is tangent to the curve y = x^2 + c, then c is...
A) 2
B) 4
C) 7
D) 11
E) 15
me and my friend got 7 and i think that's the right answer, but i don't really understand how i came to that conclusion. :/ can someone explain to me why it's C [7]?
and i do know tangent lines, so no need for any explanations about what a tangent line is. i just don't understand how c = 7 in the equation. :/
thanks!

thanks to all that contributed! our teacher never really taught us how the quadratic and the tangent lines related to each other.
i understand how we got 7 now! thanks so much! :D

Answer 1

y = x^2 + c
slope of curve at (x,y) is dy/dx = 2x
If y = 4x + 3 is tangent to the curve y = x^2 + c at (a,b) then slope of curve at (a, b) = slope of the straight line.
So 2a = 4 [since slope of st. line = 4]
or a = 2.
So, b = 4*2 + 3 = 11.
Thus the point (2, 11) lies on curve y = x^2 + c and hence must satisfy its equation.
So, 11 = 2^2 + c
or c = 7

C) is correct

Answer 2

y = 4x + 3 is tangent to the curve y = x^2 + c.

In order for this to be true, the line must cross the curve at exactly one point. This means we should obtain only one solution, should we equate these functions to each other. Let's do that.

4x + 3 = x^2 + c

Move everything to the right hand side.

0 = x^2 - 4x + c - 3

Recall that we may know the number of solutions by solving a quadratic for its discriminant, "b^2 - 4ac". If this value is 0, then we have one solution. As a reminder:

For a quadratic equation ax^2 + bx + c = 0
We can discern the nature of roots by solving b^2 - 4ac.

1) If b^2 - 4ac > 0, then we have two (unique) real solutions.
2) If b^2 - 4ac = 0, then we have one real solution (or two solutions repeated twice).
3) If b^2 - 4ac < 0, then we have two complex(imaginary) solutions.

Our goal is to make it such that #2 is true, because a line tangent to a curve crosses the curve exactly once.

Solving for "b^2 - 4ac" gives us

(-4)^2 - 4(1)(c - 3)

And we want to equate this to zero.

(-4)^2 - 4(1)(c - 3) = 0

16 - 4(c - 3) = 0

All we have to do now is solve for c.

-4(c - 3) = -16

Divide both sides by (-4),

c - 3 = 4

Add 3 to both sides,

c = 7.

Answer 3

if you solve the two equations simutaneously to get the points of intersection, you do,
x^2 + c = 4x + 3
x^2 -4x -3 +c =0
for the line to be a tangent to the curve, their should be only one point of intersection,that means this equation should have only one solution.If a quadratic equation has only one solution, that means the Discrimenant = 0
D = b^2 - 4 ac =0
a=1........b = -4..........and c= -3+c , so
(-4)^2 - 4*1* ( -3+c) =0
16 +12 - 4c = 0
28 -4c =0
4c = 28
c = 7

Answer 4

Set 4x+3 = x^2 +c
Then x^2-4x +c-3 = 0
We wish to have only one solution in order for there to be a tangency, so the discriminant must be 0
So (-4)^2 -4(1)(c-3) = 0
16 -4c+12 = 0
-4c = -28
c = 7

Answer 5

it's pretty complicated, but it is 7.

First, when you take derivative of y=x^2+c, it is 2x. from the equation of the tangent line, you know slope is 4, so 4=2x, and the point of tangency is at x=2.

Now, to the equation for point slope: y-y1=m(x-x1). when you plug in x1 as 2 and m as 4, you end up with y=4x-8+y1. because -8+y1 needs to be 3, y1 is 11.

Now we know that the point of tangency is (2,11)

now, plug (2,11) into original equation: y^2=x+c, c is 7.

Answer 6

well.. if you take the derivative of the quadratic function.. you end with
y'= 2x
now.. look at the other line.. the slope is 4.. so.
2x=4.. or x=2

if you plug this into both equations and solve for y.. you end up with 7.. try it..