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Use the quadratic formula to find the zeros of the function \(f(x)=x^{2}-4 x+13\) and then write the function in factored form. Without graphing this function, how can you tell if it intersects the \(x\) -axis?

Short Answer

Expert verified
The zeros are \(2 \pm 3i\) and the function is written as \(f(x) = (x - (2 + 3i))(x - (2 - 3i))\). It doesn't intersect the x-axis because it has complex zeros.

Step by step solution

01

Identify the coefficients

For the quadratic function \(f(x) = x^2 - 4x + 13\), identify the coefficients: a = 1 (coefficient of x^2), b = -4 (coefficient of x), c = 13 (constant term).
02

Apply the quadratic formula

The quadratic formula is \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\). Substitute the coefficients a, b, and c: \(x = \frac{-(-4) \pm \sqrt{(-4)^2 - 4(1)(13)}}{2(1)} = \frac{4 \pm \sqrt{16 - 52}}{2}\).
03

Simplify the expression under the square root

Calculate the value under the square root: \(16 - 52 = -36\). So the quadratic formula becomes \(x = \frac{4 \pm \sqrt{-36}}{2}\).
04

Simplify the square root of a negative number

Since \(-36 = 36i^2\), we have \(\sqrt{-36} = 6i\). Substitute back into the equation: \(x = \frac{4 \pm 6i}{2}\).
05

Simplify the expression

Divide each term in the numerator by 2: \(x = 2 \pm 3i\). Thus, the solutions are \(x = 2 + 3i \) and \(x = 2 - 3i\).
06

Write the function in factored form

Using the solutions, \(f(x) = (x - (2 + 3i))(x - (2 - 3i))\).
07

Determine intersection with the x-axis

Since the solutions are complex \(2 + 3i \) and \(2 - 3i\), the function has no real zeros and does not intersect the x-axis.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

quadratic formula
The quadratic formula is a powerful tool for finding the zeros of any quadratic function. It gives the solutions of a quadratic equation of the form \(ax^2 + bx + c = 0\). The formula is:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
You just need to identify the coefficients \(a\), \(b\), and \(c\) from the equation. For the function \(f(x) = x^2 - 4x + 13\), we have:
\[ a = 1, \ b = -4, \ c = 13 \]
Plugging these values into the formula gives us:
\[ x = \frac{-(-4) \pm \sqrt{(-4)^2 - 4(1)(13)}}{2(1)} \]
Simplifying, we get:
\[ x = \frac{4 \pm \sqrt{16 - 52}}{2} \]
Next, evaluate \-36\ under the square root.
complex numbers
When we have a negative number under the square root in the quadratic formula, our solutions involve complex numbers. For \-36\, we can rewrite it as \36i^2\, where \i\ is the imaginary unit with the property \i^2 = -1\. Thus, \[ \sqrt{-36} = 6i \]
Substituting back into the solution gives:
\[ x = \frac{4 \pm 6i}{2} \]
This simplifies to:
\[ x = 2 \pm 3i \]
The solutions, \2 + 3i\ and \2 - 3i\, are complex numbers.
factored form
We can express a quadratic function using its solutions. Since \ x = 2 + 3i \ and \ 2 - 3i \ are solutions, this means:
\[ (x - (2 + 3i))(x - (2 - 3i)) \]
can be used to write \ f(x) \ in its factored form. Therefore, the quadratic function becomes:
\[ f(x) = (x - (2 + 3i))(x - (2 - 3i)) \]
This representation showcases that it has complex roots.
x-axis intersection
To determine if a quadratic function intersects the \ x \-axis, we look at its solutions. If they are real, the function intersects the \ x \-axis, but if they are complex, it does not.
Since in our example \(f(x) = x^2 - 4x + 13\), the solutions \ 2 + 3i \ and \ 2 - 3i \ are complex, \ f(x) \ does not have any real solutions. Hence, the graph of \ f(x) \ does not intersect the x-axis.

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