Question

A parabola described by the equation \(y=x^2-6 x+c\) is intersected exactly once in the \(x y\)-plane by the equation \(y=-1\). What is the value of \(c\) ?

Short answer

The value of c is 8.

Step-by-step solution

Set the two equations equal to each other

Since we are looking for the intersection points where the two functions have the same y value, we can set the equations equal: \(x^2 - 6x + c = -1\)

Rearrange and simplify the equation

Rearrange the equation for x in terms of c: \(x^2 - 6x + (c + 1) = 0\)

Determine the discriminant

In a quadratic equation of the form \(ax^2 + bx + c = 0\), the discriminant (\(D\)) is given by the formula \(D = b^2 - 4ac\). In our case, \(a = 1\), \(b = -6\) and \(c = (c + 1)\). Compute the discriminant, D: \(D = (-6)^2 - 4(1)(c + 1)\)

Set the discriminant equal to 0

Since we want the parabola to intersect exactly once, this means that there should be only one real solution. This happens when the discriminant is equal to 0: \((-6)^2 - 4(1)(c + 1) = 0\)

Solve for c

Now solve the equation for c: \(36 - 4(c + 1) = 0\) \(36 - 4c - 4 = 0\) \(32 - 4c = 0\) \(4c = 32\) \(c = 8\) So, the value of c is 8.