Given the quadratic equation y=5x2+8x+9

, solve for the y-coordinate of the parabola when x=−3
.(1 point)
(−3
,
)

1 answer

To find the y-coordinate of the parabola described by the equation \( y = 5x^2 + 8x + 9 \) when \( x = -3 \), substitute \( -3 \) into the equation:

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

Calculating each term step by step:

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

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

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

  4. Add all the terms together: \[ y = 45 - 24 + 9 \] \[ y = 21 + 9 \] \[ y = 30 \]

Therefore, the y-coordinate when \( x = -3 \) is 30. The point on the parabola is \( (-3, 30) \).