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QuestionFind the zeros of the function f(x)=2x2+6x+1f(x)=2x^{2}+6x+1 using the quadratic formula. What are the x-intercepts?

Studdy Solution

STEP 1

Assumptions1. The quadratic function is given as f(x)=x+6x+1f(x)=x^{}+6x+1 . We are to find the zeros of this 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 coefficients of the quadratic equation ax+bx+c=0ax^{}+bx+c=0
4. The zeros of the function are the x-values that make the function equal to zero5. The x-intercepts of the function are the points where the graph of the function crosses the x-axis, which are also the zeros of the function

STEP 2

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

STEP 3

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

STEP 4

Substitute a=2a=2, b=6b=6, and c=1c=1 into the quadratic formula.
x=−6±62−4(2)(1)2(2)x=\frac{-6\pm\sqrt{6^{2}-4(2)(1)}}{2(2)}

STEP 5

implify the expression under the square root.
x=−±36−84x=\frac{-\pm\sqrt{36-8}}{4}

STEP 6

Calculate the value under the square root.
x=−6±284x=\frac{-6\pm\sqrt{28}}{4}

STEP 7

implify the square root.
x=−6±274x=\frac{-6\pm2\sqrt{7}}{4}

STEP 8

Divide each term in the numerator by the denominator.
x=−32±72x=-\frac{3}{2}\pm\frac{\sqrt{7}}{2}These are the zeros of the function.

STEP 9

The zeros of the function are also the x-intercepts of the graph of the function. Therefore, the x-intercepts are x=−32+72x=-\frac{3}{2}+\frac{\sqrt{7}}{2} and x=−32−72x=-\frac{3}{2}-\frac{\sqrt{7}}{2}.
So, the correct choice is A. The zeros and the x-intercepts are the same. They are x=−32+72x=-\frac{3}{2}+\frac{\sqrt{7}}{2} and x=−32−72x=-\frac{3}{2}-\frac{\sqrt{7}}{2}.

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