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Question Rewrite h(x)=2x2+11x+15h(x)=2x^2+11x+15 as (x+a)2+b(x+a)^2+b, where aa and bb are constants.

Studdy Solution

STEP 1

Assumptions
1. We are given a quadratic function h(x)=2x2+11x+15h(x) = 2x^2 + 11x + 15.
2. We need to rewrite h(x)h(x) in the form h(x)=(x+a)2+bh(x) = (x + a)^2 + b, where aa and bb are constants to be determined.
3. This process is known as completing the square.

STEP 2

To complete the square, the coefficient of x2x^2 must be 1. If it is not, we factor out the coefficient of x2x^2 from the first two terms.
h(x)=2(x2+112x)+15h(x) = 2(x^2 + \frac{11}{2}x) + 15

STEP 3

Now, we find the value to complete the square for the expression inside the parentheses. This value is (b2a)2\left(\frac{b}{2a}\right)^2, where ax2+bx+cax^2 + bx + c is the standard form of the quadratic equation.
a=1,b=112a = 1, b = \frac{11}{2}
(b2a)2=(1122)2=(114)2\left(\frac{b}{2a}\right)^2 = \left(\frac{\frac{11}{2}}{2}\right)^2 = \left(\frac{11}{4}\right)^2

STEP 4

Calculate the value to complete the square.
(114)2=(114)×(114)=12116\left(\frac{11}{4}\right)^2 = \left(\frac{11}{4}\right) \times \left(\frac{11}{4}\right) = \frac{121}{16}

STEP 5

Add and subtract the calculated value inside the parentheses to complete the square. We add and subtract 12116\frac{121}{16}, but since we factored out a 2 at the beginning, we need to multiply 12116\frac{121}{16} by 2 when subtracting it outside the parentheses to maintain equality.
h(x)=2(x2+112x+12116)+15−2×12116h(x) = 2\left(x^2 + \frac{11}{2}x + \frac{121}{16}\right) + 15 - 2 \times \frac{121}{16}

STEP 6

Simplify the expression by combining like terms.
h(x)=2(x+114)2+15−24216h(x) = 2\left(x + \frac{11}{4}\right)^2 + 15 - \frac{242}{16}

STEP 7

Convert the constant terms to have a common denominator to combine them.
15=151=15×1616=2401615 = \frac{15}{1} = \frac{15 \times 16}{16} = \frac{240}{16}
h(x)=2(x+114)2+24016−24216h(x) = 2\left(x + \frac{11}{4}\right)^2 + \frac{240}{16} - \frac{242}{16}

STEP 8

Combine the constant terms.
h(x)=2(x+114)2+(24016−24216)h(x) = 2\left(x + \frac{11}{4}\right)^2 + \left(\frac{240}{16} - \frac{242}{16}\right)
h(x)=2(x+114)2−216h(x) = 2\left(x + \frac{11}{4}\right)^2 - \frac{2}{16}

STEP 9

Simplify the constant term.
h(x)=2(x+114)2−18h(x) = 2\left(x + \frac{11}{4}\right)^2 - \frac{1}{8}

STEP 10

Finally, we rewrite the function in the desired form.
h(x)=(x+114)2−18h(x) = (x + \frac{11}{4})^2 - \frac{1}{8}
Since we want the form h(x)=(x+a)2+bh(x) = (x + a)^2 + b, we can write the solution as:
h(x)=(x+114)2−18h(x) = \left(x + \frac{11}{4}\right)^2 - \frac{1}{8}

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