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

Answer 1

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

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

Calculating step-by-step:

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

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

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

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

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

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

\[ (-3, 30) \]

Answer 2

Given this table of values, complete the table for the quadratic equation y=3x2−6x+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 (blank,blank).

Answer 3

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

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

For the equation \( y = 3x^2 - 6x + 9 \):

• \( a = 3 \)
• \( b = -6 \)

Now, substituting the values into the formula:

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

Next, we need to find the corresponding y-coordinate by substituting \( x = 1 \) back into the original equation:

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

Thus, the coordinates of the vertex (the high or low point) of the parabola are:

\[ (1, 6) \]

So the final answer is: \[ (1, 6) \]

Answer 4

Use the image to answer the question. Identify the vertex and axis of symmetry of the parabola. (1 point) axis of symmetry: x=blank vertex: (blank)

Answer 5

To identify the vertex and axis of symmetry of a parabola from a quadratic equation \(y = ax^2 + bx + c\), we can follow these steps:

1. Axis of Symmetry: The axis of symmetry is given by the formula \(x = -\frac{b}{2a}\).
2. Vertex: The vertex can be found using the x-coordinate of the axis of symmetry in the original equation to find the corresponding y-coordinate.

For a generic quadratic equation \(y = ax^2 + bx + c\):

• The x-coordinate of the vertex is \(x = -\frac{b}{2a}\).
• The y-coordinate of the vertex can be found by substituting the x-coordinate back into the equation.

Since I do not have access to images or specific values, I cannot provide exact numbers without the details about the equation or the graph. If you provide the coefficients \(a\), \(b\), and \(c\) of the quadratic equation, or describe the vertex's coordinates graphically, I can help you find the axis of symmetry and the vertex.

Answer 6

what is the answers

Answer 7

I apologize for the confusion. Without seeing the actual image or the specific parabola's equation, I can't provide the precise answers for the vertex and axis of symmetry. However, I can guide you on how to find those answers.

If you have the equation of the parabola in the form:

\[ y = ax^2 + bx + c \]

1. Axis of Symmetry: \[ x = -\frac{b}{2a} \]

2. Vertex:

   • Calculate the x-coordinate using the axis of symmetry formula.
   • Substitute the x coordinate back into the equation to find the y-coordinate.
   • The vertex will be \((x, y)\).

Example:

If your equation were \(y = 2x^2 - 4x + 1\):

• \(a = 2\), \(b = -4\)
• Axis of Symmetry: \[ x = -\frac{-4}{2 \cdot 2} = \frac{4}{4} = 1 \]
• Vertex:
  • Substitute \(x = 1\) back into the equation: \[ y = 2(1)^2 - 4(1) + 1 = 2 - 4 + 1 = -1 \]
  • So the vertex is \((1, -1)\).

Final Answers:

• Axis of Symmetry: \(x = 1\)
• Vertex: \((1, -1)\)

If you provide the specific equation or the points related to your parabola, I can give you the precise answers. Otherwise, please use the method outlined above to find the axis of symmetry and vertex for your specific equation!