Question

Given the quadratic equation y=5x2+8x+9, solve for the y-coordinate of the parabola when x=−3.(1 point) (−3 ,)

Answer 1

To find the y-coordinate of the parabola defined by the quadratic equation \( y = 5x^2 + 8x + 9 \) when \( x = -3 \), substitute \( -3 \) for \( x \) in 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 \( 8(-3) \):
\[
8 \times -3 = -24
\]

Now substitute these values back into the equation:
\[
y = 45 - 24 + 9
\]

Combine the values:
\[
y = 45 - 24 = 21
\]
\[
y = 21 + 9 = 30
\]

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

Thus, the coordinates are \( (-3, 30) \).