Solved on Jan 18, 2024

Rewrite the equation $y=2x^2+8x+9$ in vertex form by completing the square, then graph the resulting function.

STEP 1

Assumptions
1. The given quadratic function is $y = 2x^2 + 8x + 9$ .
2. We want to write the function in vertex form, which is $y = a(x-h)^2 + k$ , where $(h, k)$ is the vertex of the parabola.
3. Completing the square is a method used to convert a quadratic function into vertex form.

STEP 2

To complete the square, we need to create a perfect square trinomial from the quadratic and linear terms of the given function. The constant term is temporarily set aside during this process.
$y = 2x^2 + 8x + 9$

STEP 3

Factor out the coefficient of the $x^2$ term from the first two terms.
$y = 2(x^2 + 4x) + 9$

STEP 4

To complete the square inside the parentheses, we need to find a number that, when added and then factored, creates a perfect square trinomial. This number is $\left(\frac{b}{2a}\right)^2$ , where $a$ is the coefficient of $x^2$ and $b$ is the coefficient of $x$ .
$\left(\frac{b}{2a}\right)^2 = \left(\frac{4}{2 \cdot 1}\right)^2 = (2)^2 = 4$

STEP 5

Add and subtract the number found in STEP_4 inside the parentheses. We subtract it to maintain the equality because we are effectively adding zero.
$y = 2(x^2 + 4x + 4 - 4) + 9$

STEP 6

Simplify the expression by combining the constant terms outside the parentheses.
$y = 2(x^2 + 4x + 4) - 2 \cdot 4 + 9$

STEP 7

Factor the perfect square trinomial inside the parentheses.
$y = 2(x + 2)^2 - 8 + 9$

STEP 8

Combine the constant terms to complete the vertex form.
$y = 2(x + 2)^2 + 1$

STEP 9

Now we have the function in vertex form. The vertex $(h, k)$ of the parabola can be read directly from the equation as $(-2, 1)$ .

STEP 10

To graph the function, we start by plotting the vertex $(-2, 1)$ on the coordinate plane.

STEP 11

Since the coefficient of the $(x + 2)^2$ term is positive (2), we know that the parabola opens upwards.

STEP 12

The next step is to find additional points on the parabola by choosing $x$ -values around the vertex and plugging them into the vertex form equation to find the corresponding $y$ -values.

STEP 13

Plot the points found in STEP_12 on the graph and draw a smooth curve through the vertex and these points to complete the graph of the parabola.
The vertex form of the given quadratic function is $y = 2(x + 2)^2 + 1$ , and the vertex of the parabola is $(-2, 1)$ .

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Rewrite the equation $y=2x^2+8x+9$ in vertex form by completing the square, then graph the resulting function.

STEP 1

Assumptions
1. The given quadratic function is $y = 2x^2 + 8x + 9$ .
2. We want to write the function in vertex form, which is $y = a(x-h)^2 + k$ , where $(h, k)$ is the vertex of the parabola.
3. Completing the square is a method used to convert a quadratic function into vertex form.

STEP 2

To complete the square, we need to create a perfect square trinomial from the quadratic and linear terms of the given function. The constant term is temporarily set aside during this process.
$y = 2x^2 + 8x + 9$

STEP 3

Factor out the coefficient of the $x^2$ term from the first two terms.
$y = 2(x^2 + 4x) + 9$

STEP 4

STEP 5

Add and subtract the number found in STEP_4 inside the parentheses. We subtract it to maintain the equality because we are effectively adding zero.
$y = 2(x^2 + 4x + 4 - 4) + 9$

STEP 6

Simplify the expression by combining the constant terms outside the parentheses.
$y = 2(x^2 + 4x + 4) - 2 \cdot 4 + 9$

STEP 7

Factor the perfect square trinomial inside the parentheses.
$y = 2(x + 2)^2 - 8 + 9$

STEP 8

Combine the constant terms to complete the vertex form.
$y = 2(x + 2)^2 + 1$

STEP 9

Now we have the function in vertex form. The vertex $(h, k)$ of the parabola can be read directly from the equation as $(-2, 1)$ .

STEP 10

To graph the function, we start by plotting the vertex $(-2, 1)$ on the coordinate plane.

STEP 11

Since the coefficient of the $(x + 2)^2$ term is positive (2), we know that the parabola opens upwards.

STEP 12

The next step is to find additional points on the parabola by choosing $x$ -values around the vertex and plugging them into the vertex form equation to find the corresponding $y$ -values.

STEP 13

Plot the points found in STEP_12 on the graph and draw a smooth curve through the vertex and these points to complete the graph of the parabola.
The vertex form of the given quadratic function is $y = 2(x + 2)^2 + 1$ , and the vertex of the parabola is $(-2, 1)$ .

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Rewrite the equation y=2x2+8x+9y=2x^2+8x+9y=2x2+8x+9 in vertex form by completing the square, then graph the resulting function.

STEP 1

STEP 2

To complete the square, we need to create a perfect square trinomial from the quadratic and linear terms of the given function. The constant term is temporarily set aside during this process.y=2x2+8x+9y = 2x^2 + 8x + 9y=2x2+8x+9

STEP 3

Factor out the coefficient of the x2x^2x2 term from the first two terms.y=2(x2+4x)+9y = 2(x^2 + 4x) + 9y=2(x2+4x)+9

STEP 4

STEP 5

Add and subtract the number found in STEP_4 inside the parentheses. We subtract it to maintain the equality because we are effectively adding zero.y=2(x2+4x+4−4)+9y = 2(x^2 + 4x + 4 - 4) + 9y=2(x2+4x+4−4)+9

STEP 6

Simplify the expression by combining the constant terms outside the parentheses.y=2(x2+4x+4)−2⋅4+9y = 2(x^2 + 4x + 4) - 2 \cdot 4 + 9y=2(x2+4x+4)−2⋅4+9

STEP 7

Factor the perfect square trinomial inside the parentheses.y=2(x+2)2−8+9y = 2(x + 2)^2 - 8 + 9y=2(x+2)2−8+9

STEP 8

Combine the constant terms to complete the vertex form.y=2(x+2)2+1y = 2(x + 2)^2 + 1y=2(x+2)2+1

STEP 9

Now we have the function in vertex form. The vertex (h,k)(h, k)(h,k) of the parabola can be read directly from the equation as (−2,1)(-2, 1)(−2,1).

STEP 10

To graph the function, we start by plotting the vertex (−2,1)(-2, 1)(−2,1) on the coordinate plane.

STEP 11

Since the coefficient of the (x+2)2(x + 2)^2(x+2)2 term is positive (2), we know that the parabola opens upwards.

STEP 12

The next step is to find additional points on the parabola by choosing xxx-values around the vertex and plugging them into the vertex form equation to find the corresponding yyy-values.

STEP 13

Plot the points found in STEP_12 on the graph and draw a smooth curve through the vertex and these points to complete the graph of the parabola.The vertex form of the given quadratic function is y=2(x+2)2+1y = 2(x + 2)^2 + 1y=2(x+2)2+1, and the vertex of the parabola is (−2,1)(-2, 1)(−2,1).

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Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

Rewrite the equation $y=2x^2+8x+9$ in vertex form by completing the square, then graph the resulting function.

To complete the square, we need to create a perfect square trinomial from the quadratic and linear terms of the given function. The constant term is temporarily set aside during this process.
$y = 2x^2 + 8x + 9$

Factor out the coefficient of the $x^2$ term from the first two terms.
$y = 2(x^2 + 4x) + 9$

Add and subtract the number found in STEP_4 inside the parentheses. We subtract it to maintain the equality because we are effectively adding zero.
$y = 2(x^2 + 4x + 4 - 4) + 9$

Simplify the expression by combining the constant terms outside the parentheses.
$y = 2(x^2 + 4x + 4) - 2 \cdot 4 + 9$

Factor the perfect square trinomial inside the parentheses.
$y = 2(x + 2)^2 - 8 + 9$

Combine the constant terms to complete the vertex form.
$y = 2(x + 2)^2 + 1$

Now we have the function in vertex form. The vertex $(h, k)$ of the parabola can be read directly from the equation as $(-2, 1)$ .

To graph the function, we start by plotting the vertex $(-2, 1)$ on the coordinate plane.

Since the coefficient of the $(x + 2)^2$ term is positive (2), we know that the parabola opens upwards.

The next step is to find additional points on the parabola by choosing $x$ -values around the vertex and plugging them into the vertex form equation to find the corresponding $y$ -values.

Plot the points found in STEP_12 on the graph and draw a smooth curve through the vertex and these points to complete the graph of the parabola.
The vertex form of the given quadratic function is $y = 2(x + 2)^2 + 1$ , and the vertex of the parabola is $(-2, 1)$ .