Question

The line y = 4x − 7 is tangent to a parabola that has a y-intercept of −3 and the line x = 1/2
as its axis of symmetry. Find the equation of the parabola.

Answer 1

@oldrin.bataku help please

Answer 2

id start by forming the vertex form of the parabola ....

Answer 3

ill type what i got so far :P
y = ax^2 + bx - 3
x=1/2
y = a/4 + b/2 - 3

Answer 4

hmm

x = -b/2a = 1/2; b=-k, a=k would fit for generality

y = kx^2 - kx -3

Answer 5

with the parabola observe \(y=a(x-1/2)^2+k\). for a y-intercept of \(-3\) we want:$$y(0)=-3\\a(-1/2)^2+k=-3\\a/4+k=-3\\k=-a/4-3$$hence so far we have \(y=a(x-1/2)^2-a/4-3\).
Okay, now since \(y=4x-7\) is tangent we expect them to intersect at one point:$$a(x-1/2)^2-a/4-3=4x-7\\a(x^2-x+1/2)-a/2-3=4x-7\\ax^2-ax-3=4x-7\\ax^2-(a+4)x+4=0$$

Answer 6

now, since we want there to be only one point of intersection, the above quadratic must have one and only one real root; this corresponds to our discriminant being \(0\):$$(a+4)^2-16a=0\\a^2+8a+16-16a=0\\a^2-8a+16=0\\(a-4)^2=0\\a=4$$hence our quadratic equation is just:$$y=4(x-1/2)^2-1-3=4(x-1/2)^2-4$$

Answer 7

wolframalpha suggests my approach was correct: http://www.wolframalpha.com/input/?i=y%3D4%28x-1%2F2%29%5E2-4%2C+y%3D4x-7

Answer 8

a(−1/2)^2+k=−3
how did you get this?

Answer 9

@AonZ vertex form of a quadratic equation, \(y=a(x-h)^2+k\) where \(x=h\) is our axis of symmetry.

Answer 10

@AonZ recall the y-intercept is just the value of \(y\) when \(x=0\) so plug in \(x=0\) into \(y=a(x-1/2)^2+k\) and we get:$$y=a(0-1/2)^2+k=a(-1/2)^2+k=a/4+k$$since we want our y-intercept to be \(-3\) we want \(y=-3\) for \(x=0\) hence:$$-3=a/4+k\\k=-a/4-3$$

Answer 11

ahh ok thanks :)

Answer 12

the discriminant being \(0\) is from the quadratic formula, btw; for \(ax^2+bx+c=0\) to have one real solution observe:$$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$$that we need \(b^2-4ac=0\) in which case the above reduces to \(x=-\dfrac{b}{2a}\) (since a vertical parabola can only touch the \(x\) axis at only 1 point if it touches at its vertex)

Answer 13

Ok, thanks for all your help :)