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QuestionFind the real zeros of f(x)=x2+6x+4f(x)=x^{2}+6x+4 using the quadratic formula. What are the xx-intercepts?

Studdy Solution

STEP 1

Assumptions1. The function is a quadratic function given by f(x)=x+6x+4f(x)=x^{}+6 x+4 . We are asked to find the real zeros of the function using the quadratic formula3. The quadratic formula is given by x=−b±b−4acax = \frac{-b \pm \sqrt{b^ -4ac}}{a}, where aa, bb, and cc are the coefficients of the quadratic equation ax+bx+c=0ax^ + bx + c =0
4. The x-intercepts of a function are the x-values where the function equals zero, i.e., f(x)=0f(x) =0

STEP 2

First, we need to identify the coefficients aa, bb, and cc from the quadratic function f(x)=x2+6x+4f(x)=x^{2}+6 x+4.
a=1,b=6,c=4a =1, b =6, c =4

STEP 3

Now, we can plug these values into the quadratic formula to find the solutions for xx.
x=−b±b2−ac2ax = \frac{-b \pm \sqrt{b^2 -ac}}{2a}x=−6±62−∗1∗2∗1x = \frac{-6 \pm \sqrt{6^2 -*1*}}{2*1}

STEP 4

implify the expression under the square root.
x=−6±36−162x = \frac{-6 \pm \sqrt{36 -16}}{2}

STEP 5

Calculate the value under the square root.
x=−±202x = \frac{- \pm \sqrt{20}}{2}

STEP 6

implify the square root.
x=−6±252x = \frac{-6 \pm2\sqrt{5}}{2}

STEP 7

implify the expression to find the solutions for xx.
x=−3±5x = -3 \pm \sqrt{5}The real zeros of the function are x=−3+5x = -3 + \sqrt{5} and x=−3−5x = -3 - \sqrt{5}.

STEP 8

The x-intercepts of a function are the x-values where the function equals zero. Since the real zeros of a function are the solutions to the equation f(x)=0f(x) =0, the x-intercepts and the real zeros are the same for any given function.
Therefore, the x-intercepts of the function are x=−3+5x = -3 + \sqrt{5} and x=−3−5x = -3 - \sqrt{5}.
The correct choice isA. The zeros and the xx-intercepts are the same. They are x=−3+5x = -3 + \sqrt{5} and x=−3−5x = -3 - \sqrt{5}.

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