Given the quadratic equation y=5x^2+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 given the equation \( y = 5x^2 + 8x + 9 \) when \( x = -3 \), we substitute \( -3 \) into the equation for \( x \):

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

Now, calculate each term:

1. \( (-3)^2 = 9 \) \[ 5 \times 9 = 45 \]

2. \( 8 \times (-3) = -24 \)

Now, add these results together along with 9:

\[ y = 45 - 24 + 9 \]

Calculating further:

\[ y = 45 - 24 = 21 \] \[ y = 21 + 9 = 30 \]

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

So, the point is \( (-3, 30) \).

Answer 2

Given this table of values, complete the table for the quadratic equation y=3x^2−6x+9 to find the values of the high or low point of the parabola. x y −3 54 −2 33 −1 0 1 2 9 (1 point) The high or low point of this parabola is ((),()).

Answer 3

To find the high or low point (vertex) of the parabola given by the equation \( y = 3x^2 - 6x + 9 \), we first need to identify the vertex using the formula for the x-coordinate of the vertex of a quadratic equation in the standard form \( ax^2 + bx + c \):

\[ x = -\frac{b}{2a} \]

In this case:

• \( a = 3 \)
• \( b = -6 \)

Substituting these values into the formula gives:

\[ x = -\frac{-6}{2 \cdot 3} = \frac{6}{6} = 1 \]

Next, we will calculate the corresponding y-coordinate by substituting \( x = 1 \) into the equation:

\[ y = 3(1)^2 - 6(1) + 9 \]

Calculating step-by-step:

1. \( 3(1)^2 = 3 \)
2. \( -6(1) = -6 \)

Now combine these:

\[ y = 3 - 6 + 9 = 6 \]

Thus, the high or low point (vertex) of the parabola is at the point \( (1, 6) \).

Therefore, the completion of the statement is:

The high or low point of this parabola is \( (1, 6) \).