Graphs of Quadratic Equations Practice

Question

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

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

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

Now, calculate 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. Now put it all together: \[ y = 45 - 24 + 9 \]

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

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

The point is: \[ (-3, 30) \]