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QuestionFind the real zeros of the function f(x)=x2+4x+1f(x)=x^{2}+4x+1 using the quadratic formula. What are the x-intercepts?

Studdy Solution

STEP 1

Assumptions1. The quadratic function is given as f(x)=x+4x+1f(x)=x^{}+4 x+1 . We are to find the real zeros of the function using the quadratic formula3. The quadratic formula is given as x=−b±b−4acax = \frac{-b \pm \sqrt{b^ -4ac}}{a} for a quadratic equation ax+bx+c=0ax^ + bx + c =0
4. The real zeros of the function are also the x-intercepts of the graph of the function

STEP 2

First, we need to identify the coefficients aa, bb, and cc from the given quadratic function.
In our case, a=1a=1, b=4b=4, and c=1c=1.

STEP 3

Now, we can substitute these values into the quadratic formula.
x=−b±b2−ac2ax = \frac{-b \pm \sqrt{b^2 -ac}}{2a}

STEP 4

Substitute a=1a=1, b=4b=4, and c=1c=1 into the formula.
x=−4±42−4∗1∗12∗1x = \frac{-4 \pm \sqrt{4^2 -4*1*1}}{2*1}

STEP 5

implify the expression under the square root.
x=−4±16−42x = \frac{-4 \pm \sqrt{16 -4}}{2}

STEP 6

Calculate the value under the square root.
x=−4±122x = \frac{-4 \pm \sqrt{12}}{2}

STEP 7

implify the square root.
x=−4±232x = \frac{-4 \pm2\sqrt{3}}{2}

STEP 8

implify the expression by dividing each term by2.
x=−2±3x = -2 \pm \sqrt{3}These are the real zeros of the function.

STEP 9

The x-intercepts of the graph of the function are the same as the real zeros of the function. Therefore, the x-intercepts are also −2±3-2 \pm \sqrt{3}.
So, the correct choice is A. The zeros and the x-intercepts are the same. They are −2+3-2 + \sqrt{3} and −2−3-2 - \sqrt{3}.

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