Question

Find the minimum value of the parabola
y

=
x
2

−
8
x
+
31
5
.


Simplify your answer and write it as a proper fraction, improper fraction, or integer.

Answer 1

To find the minimum value of the parabola given by the equation

\[
y = x^2 - 8x + \frac{31}{5},
\]

we can complete the square.

1. Start by focusing on the quadratic part \(x^2 - 8x\). To complete the square, we take half of the coefficient of \(x\), square it, and adjust the expression. The coefficient of \(x\) is \(-8\):

\[
\left(-\frac{8}{2}\right)^2 = (-4)^2 = 16.
\]

2. Rewrite the quadratic part using this square:

\[
x^2 - 8x = (x - 4)^2 - 16.
\]

3. Substitute back into the original equation:

\[
y = (x - 4)^2 - 16 + \frac{31}{5}.
\]

4. To combine \(-16\) and \(\frac{31}{5}\), convert \(-16\) into a fraction with a denominator of 5:

\[
-16 = -\frac{80}{5}.
\]

5. Now substitute this into the equation:

\[
y = (x - 4)^2 - \frac{80}{5} + \frac{31}{5} = (x - 4)^2 - \frac{80 - 31}{5} = (x - 4)^2 - \frac{49}{5}.
\]

6. The expression \( (x - 4)^2 \) reaches its minimum value of 0 when \(x = 4\). Therefore, the minimum value of \(y\) occurs at:

\[
y = 0 - \frac{49}{5} = -\frac{49}{5}.
\]

Thus, the minimum value of the parabola is

\[
\boxed{-\frac{49}{5}}.
\]