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QuestionFind the real zeros of the function f(x)=x2+8x+14f(x)=x^{2}+8x+14 using the quadratic formula. What are the x-intercepts?

Studdy Solution

STEP 1

Assumptions1. The quadratic function is given as f(x)=x+8x+14f(x) = x^{} +8x +14. . We are to find the real zeros of this function using the quadratic formula.
3. The quadratic formula is given as x=−b±b−4acax = \frac{-b \pm \sqrt{b^ -4ac}}{a}, where aa, bb, and cc are coefficients of the quadratic equation in the form ax+bx+cax^ + bx + c.
4. The real zeros of the function are the same as the x-intercepts of the graph of the function.

STEP 2

First, identify the coefficients aa, bb, and cc from the given quadratic function. In f(x)=x2+8x+14f(x) = x^{2} +8x +14, we havea=1a =1, b=8b =8, and c=14c =14.

STEP 3

Next, substitute these coefficients into the quadratic formula.x=−b±b2−ac2ax = \frac{-b \pm \sqrt{b^2 -ac}}{2a}becomesx=−8±82−∗1∗142∗1x = \frac{-8 \pm \sqrt{8^2 -*1*14}}{2*1}

STEP 4

implify the expression under the square root, 82−4∗1∗148^2 -4*1*14.
x=−8±64−562x = \frac{-8 \pm \sqrt{64 -56}}{2}

STEP 5

Further simplify the expression under the square root, 64−5664 -56.
x=−8±82x = \frac{-8 \pm \sqrt{8}}{2}

STEP 6

Now, simplify the square root, 8\sqrt{8}.
x=−8±222x = \frac{-8 \pm2\sqrt{2}}{2}

STEP 7

Finally, simplify the expression by dividing each term in the numerator by2.
x=−4±2x = -4 \pm \sqrt{2}The real zeros of the function are x=−4+2x = -4 + \sqrt{2} and x=−4−2x = -4 - \sqrt{2}.

STEP 8

The x-intercepts of the graph of a function are the x-values for which f(x)=0f(x) =0. In other words, the x-intercepts are the real zeros of the function. Therefore, the x-intercepts of the graph of the function are also x=−4+2x = -4 + \sqrt{2} and x=−4−2x = -4 - \sqrt{2}.
So, the correct choice is B. The zeros and the x-intercepts are the same. They are x=−4+2x = -4 + \sqrt{2} and x=−4−2x = -4 - \sqrt{2}.

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