Problem 176 - Elementary Algebra

Question

In the following exercises, find the \(x\) - and \(y\) -intercepts. $$ y=x^{2}+6 x+13 $$

Step-by-Step Solution

Answer

The y-intercept is \((0, 13)\) and the x-intercepts are complex: \(x = -3 \pm 2i\).

Step 1: Find the y-intercept

To find the y-intercept, set \(x = 0\) and solve for \(y\). Substitute \(x = 0\) into \(y = x^2 + 6x + 13\), resulting in: \(y = 0^2 + 6(0) + 13 = 13\). So, the y-intercept is \(13\), or the point \((0, 13)\).

Step 2: Find the x-intercepts

To find the x-intercepts, set \(y = 0\) and solve for \(x\). So, \(0 = x^2 + 6x + 13\). This equation can be solved using the quadratic formula \(x = \frac {-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 1\), \(b = 6\), and \(c = 13\).

Step 3: Apply the quadratic formula

Substitute the values of \(a\), \(b\), and \(c\) into the quadratic formula: \(x = \frac {-6 \pm \sqrt{6^2 - 4(1)(13)}}{2(1)} = \frac {-6 \pm \sqrt{36 - 52}}{2}\). This simplifies to: \(x = \frac {-6 \pm \sqrt{-16}}{2}\).

Step 4: Simplify the solution

Since \(\sqrt{-16} = 4i\) (where \(i\) is the imaginary unit), the equation becomes: \(x = \frac {-6 \pm 4i}{2} = -3 \pm 2i\). Thus, the x-intercepts are complex numbers: \(x = -3 + 2i\) and \(x = -3 - 2i\).

Key Concepts

y-intercept

The y-intercept of a function is where the graph crosses the y-axis. To find it, set the value of x to zero and solve for y. For example, with the function \( y = x^2 + 6x + 13 \), we substitute x with 0. This gives us:

• \( y = 0^2 + 6(0) + 13 \)
• \( y = 13 \)

So, the y-intercept is at the point (0, 13). This tells us that when x is zero, y equals 13. This point is important because it helps us understand where the graph meets the y-axis.

x-intercepts

The x-intercepts are the points where the graph crosses the x-axis. To find them, set y to zero and solve for x. For our example, we start with:

• \( 0 = x^2 + 6x + 13 \)

Now, we need to solve this quadratic equation. The solution might involve real numbers or complex numbers, depending on the discriminant \( b^2 - 4ac \). Let's use the quadratic formula to find the x-intercepts in more detail.

quadratic formula

The quadratic formula is used to solve quadratic equations of the form \( ax^2 + bx + c = 0 \). The formula is given by: \[ x = \frac {-b \pm \sqrt{b^2 - 4ac}}{2a} \] In our case, \( a = 1 \), \( b = 6 \), and \( c = 13 \). Substituting these into the formula, we get: \[ x = \frac {-6 \pm \sqrt{6^2 - 4(1)(13)}}{2(1)} = \frac {-6 \pm \sqrt{36 - 52}}{2} \] This simplifies to: \[ x = \frac {-6 \pm \sqrt{-16}}{2} \] Because the expression under the square root is negative, the solutions involve complex numbers.

complex numbers

Complex numbers arise when we take the square root of a negative number. The imaginary unit \( i \) is defined as \( \sqrt{-1} \). Thus, \( \sqrt{-16} = 4i \). Using this in our quadratic formula, we have:

• \( x = \frac {-6 \pm 4i}{2} = -3 \pm 2i \)

So, the x-intercepts are actually complex numbers: \( -3 + 2i \) and \( -3 - 2i \). These points do not cross the x-axis on a real number plane, but they are crucial in understanding the complete behavior of quadratic functions, including those that do not have real-number solutions. Complex numbers are essential for solving equations where real numbers are not sufficient.