Is (–8, 0) a solution to the equation y = x2 − 5?

1 answer

To determine if the point \((-8, 0)\) is a solution to the equation \(y = x^2 - 5\), we can substitute \(x = -8\) into the equation and see if we get \(y = 0\).

Starting with the equation:

\[ y = x^2 - 5 \]

Substituting \(x = -8\):

\[ y = (-8)^2 - 5 \]

Calculating \((-8)^2\):

\[ (-8)^2 = 64 \]

Now substituting back into the equation:

\[ y = 64 - 5 = 59 \]

Now we have \(y = 59\) when \(x = -8\). The point \((-8, 0)\) gives \(y = 0\), but we've determined that \(y = 59\).

Since \(0 \neq 59\), the point \((-8, 0)\) is not a solution to the equation \(y = x^2 - 5\).