Find the Bounds of the Zeros f(x)=2x^2+6

$f (x) = 2 x^{2} + 6$

Step 1

Check the leading coefficient of the function. This number is the coefficient of the expression with the largest degree.

Largest Degree: $2$

Leading Coefficient: $2$

Step 2

Simplify each term.

Cancel the common factor of $2$ .

Cancel the common factor.

$f (x) = \frac{2 x^{2}}{2} + \frac{6}{2}$

Divide $x^{2}$ by $1$ .

$f (x) = x^{2} + \frac{6}{2}$

Divide $6$ by $2$ .

$f (x) = x^{2} + 3$

Step 3

Create a list of the coefficients of the function except the leading coefficient of $1$ .

$3$

Step 4

There will be two bound options, $b_{1}$ and $b_{2}$ , the smaller of which is the answer. To calculate the first bound option, find the absolute value of the largest coefficient from the list of coefficients. Then add $1$ .

Arrange the terms in ascending order.

$b_{1} = | 3 |$

The absolute value is the distance between a number and zero. The distance between $0$ and $3$ is $3$ .

$b_{1} = 3 + 1$

Add $3$ and $1$ .

$b_{1} = 4$

Step 5

To calculate the second bound option, sum the absolute values of the coefficients from the list of coefficients. If the sum is greater than $1$ , use that number. If not, use $1$ .

The absolute value is the distance between a number and zero. The distance between $0$ and $3$ is $3$ .

$b_{2} = 3$

Arrange the terms in ascending order.

$b_{2} = 1, 3$

The maximum value is the largest value in the arranged data set.

$b_{2} = 3$

Step 6

Take the smaller bound option between $b_{1} = 4$ and $b_{2} = 3$ .

Smaller Bound: $3$

Step 7

Every real root on $f (x) = 2 x^{2} + 6$ lies between $- 3$ and $3$ .

$- 3$ and $3$

Find the Bounds of the Zeros f(x)=2x^2+6

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