CALCU-how to do 4a)4 a Express 2 x^2+9 x+4 in the

how to do 4a)

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Solution by Steps

step 1

To express

2x^2 + 9x + 4

in the form

a(x+b)^2 + c

, we start by completing the square

step 2

Factor out the coefficient of

x^2

from the first two terms:

2(x^2 + \frac{9}{2}x) + 4

step 3

To complete the square inside the parentheses, add and subtract

(\frac{9}{4})^2

:

2(x^2 + \frac{9}{2}x + \frac{81}{16} - \frac{81}{16}) + 4

step 4

Simplify inside the parentheses:

2((x + \frac{9}{4})^2 - \frac{81}{16}) + 4

step 5

Distribute the 2 and combine constants:

2(x + \frac{9}{4})^2 - \frac{81}{8} + 4

step 6

Combine the constants:

2(x + \frac{9}{4})^2 - \frac{49}{8}

Answer

2(x + \frac{9}{4})^2 - \frac{49}{8}

Key Concept

Completing the Square

Explanation

Completing the square involves rewriting a quadratic expression in the form

a(x+b)^2 + c

by adding and subtracting the same value inside the parentheses.

Solution by Steps

step 1

To find the vertex of the parabola

y = 2x^2 + 9x + 4

, we use the completed square form

y = 2(x + \frac{9}{4})^2 - \frac{49}{8}

step 2

The vertex form of a parabola

y = a(x-h)^2 + k

gives the vertex at

(h, k)

step 3

From

y = 2(x + \frac{9}{4})^2 - \frac{49}{8}

, the vertex is

(-\frac{9}{4}, -\frac{49}{8})

step 4

Since the coefficient of

(x + \frac{9}{4})^2

is positive (2), the parabola opens upwards, indicating a minimum point

Answer

The vertex is

(-\frac{9}{4}, -\frac{49}{8})

and it is a minimum point.

Key Concept

Vertex of a Parabola

Explanation

The vertex of a parabola in the form

y = a(x-h)^2 + k

is

(h, k)

. If a > 0, the vertex is a minimum point; if a < 0, it is a maximum point.

how to do 6b)

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Solution by Steps

step 1

To write

1 + x - 2x^2

in the form

p - 2(x - q)^2

, we start by completing the square

step 2

Rewrite the expression

1 + x - 2x^2

as

-2x^2 + x + 1

step 3

Factor out

-2

from the quadratic and linear terms:

-2(x^2 - \frac{1}{2}x) + 1

step 4

To complete the square inside the parentheses, add and subtract

(\frac{1}{4})^2

:

-2(x^2 - \frac{1}{2}x + \frac{1}{16} - \frac{1}{16}) + 1

step 5

Simplify the expression:

-2((x - \frac{1}{4})^2 - \frac{1}{16}) + 1

step 6

Distribute

-2

:

-2(x - \frac{1}{4})^2 + \frac{2}{16} + 1

step 7

Combine constants:

-2(x - \frac{1}{4})^2 + \frac{1}{8} + 1 = -2(x - \frac{1}{4})^2 + \frac{9}{8}

step 8

Therefore,

1 + x - 2x^2

can be written as

p - 2(x - q)^2

where

p = \frac{9}{8}

and

q = \frac{1}{4}

Answer

1 + x - 2x^2 = \frac{9}{8} - 2(x - \frac{1}{4})^2

Key Concept

Completing the Square

Explanation

Completing the square is a method used to convert a quadratic expression into a perfect square trinomial plus or minus a constant. This form is useful for identifying the vertex of a parabola.

Solution by Steps

step 1

To sketch the graph of

y = 1 + x - 2x^2

, we first note that it is a downward-opening parabola because the coefficient of

x^2

is negative

step 2

The vertex form of the equation is

y = \frac{9}{8} - 2(x - \frac{1}{4})^2

step 3

The vertex of the parabola is at

(\frac{1}{4}, \frac{9}{8})

step 4

The parabola opens downwards, so it has a maximum point at the vertex

step 5

To sketch the graph, plot the vertex at

(\frac{1}{4}, \frac{9}{8})

and draw a symmetric parabola opening downwards

Answer

The graph of

y = 1 + x - 2x^2

is a downward-opening parabola with vertex at

(\frac{1}{4}, \frac{9}{8})

.

Key Concept

Graphing Parabolas

Explanation

The vertex form of a quadratic equation helps in easily identifying the vertex and the direction in which the parabola opens, which is crucial for sketching its graph.

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