Given the quadratic equation y = 5x^2 + 8x + 9, solve for the y-coordinate of the parabola when x = −3.

(-3, ___)

1 answer

To find the y-coordinate of the parabola described by the equation \( y = 5x^2 + 8x + 9 \) when \( x = -3 \), we can substitute \( -3 \) for \( x \) in the equation.

Starting with the equation:

\[ y = 5x^2 + 8x + 9 \]

Substituting \( x = -3 \):

\[ y = 5(-3)^2 + 8(-3) + 9 \]

Now, calculating each term step-by-step:

  1. Calculate \( (-3)^2 \): \[ (-3)^2 = 9 \]

  2. Multiply by 5: \[ 5 \cdot 9 = 45 \]

  3. Calculate \( 8(-3) \): \[ 8 \cdot (-3) = -24 \]

Now substituting these back into the equation for \( y \):

\[ y = 45 - 24 + 9 \]

Calculating it step-by-step:

  1. First combine \( 45 - 24 \): \[ 45 - 24 = 21 \]

  2. Now add \( 9 \): \[ 21 + 9 = 30 \]

Thus, the y-coordinate when \( x = -3 \) is:

\[ y = 30 \]

So the point on the parabola is:

\[ (-3, 30) \]