Problem 36 - Elementary Algebra

Question

\(y=x^{2}-8 x+10\)

Step-by-Step Solution

Answer

The vertex is (4, -6) and the y-intercept is (0, 10). The x-intercepts are \(4 + \sqrt{6}\) and \(4 - \sqrt{6}\).

Step 1: Identify the Quadratic Equation

Recognize that the given equation is a quadratic equation: \(y = x^2 - 8x + 10\).

Step 2: Write in Standard Form

The equation is already in standard form, which is \(ax^2 + bx + c\) where \(a = 1\), \(b = -8\), and \(c = 10\).

Step 3: Find the Vertex

Use the vertex formula \(x = -\frac{b}{2a}\). Here, \(a = 1\) and \(b = -8\), so \(x = -\frac{-8}{2(1)} = 4\). To find the y-coordinate of the vertex, plug \(x = 4\) back into the equation: \(y = (4)^2 - 8(4) + 10 = 16 - 32 + 10 = -6\). The vertex is (4, -6).

Step 4: Determine the Axis of Symmetry

The axis of symmetry is the vertical line that passes through the vertex. Thus, the axis of symmetry is \(x = 4\).

Step 5: Find the Y-Intercept

Substitute \(x = 0\) into the equation to find the y-intercept: \(y = (0)^2 - 8(0) + 10 = 10\). Therefore, the y-intercept is (0, 10).

Step 6: Find the X-Intercepts (if they exist)

Set \(y = 0\) and solve the quadratic equation \(0 = x^2 - 8x + 10\). Use the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\). Here, \(a = 1\), \(b = -8\), and \(c = 10\). Calculate the discriminant: \(b^2 - 4ac = (-8)^2 - 4(1)(10) = 64 - 40 = 24\). Since the discriminant is positive, there are two real solutions: \(x = \frac{8 \pm \sqrt{24}}{2} = 4 \pm \sqrt{6}\). The x-intercepts are \(4 + \sqrt{6}\) and \(4 - \sqrt{6}\).

Key Concepts

Quadratic Formula

The quadratic formula is a powerful tool for solving quadratic equations of the form \(ax^2 + bx + c = 0\). If you have trouble factoring the equation or if it does not factor neatly, the quadratic formula can be very useful.
The formula is: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]Here’s how it works:

• Identify the coefficients \(a\), \(b\), and \(c\) from the quadratic equation.
• Calculate the discriminant, \(b^2 - 4ac\), inside the square root.
• Use the values of \(a\), \(b\), and the discriminant in the quadratic formula to find the solutions for \(x\).

In our exercise, the discriminant was 24, leading to two real and distinct solutions. This tells us where the parabola intersects the \(x\)-axis.

Vertex of a Parabola

The vertex of a parabola represents its highest or lowest point, depending on the parabola's orientation. For the quadratic function \(y = ax^2 + bx + c\):

• The x-coordinate of the vertex can be found using \(x = -\frac{b}{2a}\).
• The y-coordinate is found by substituting the x-coordinate back into the original equation.

In the given quadratic equation \(y = x^2 - 8x + 10\), we calculated the vertex to be at \(x = 4\). After substituting \(x = 4\) back into the equation, we found that the y-coordinate is -6. Thus, the vertex is (4, -6). This represents the lowest point of this parabola.

Axis of Symmetry

The axis of symmetry of a parabola is a vertical line that divides the parabola into two symmetrical halves. It always passes through the vertex. For a quadratic equation \(y = ax^2 + bx + c\), the axis of symmetry can be determined by:

• Using the x-coordinate of the vertex \(x = -\frac{b}{2a}\).

In our exercise, since the vertex is at (4, -6), the axis of symmetry is the line \(x = 4\). This means that the parabola is symmetrical on either side of the line \(x = 4\).

Y-Intercept

The y-intercept is the point at which the parabola intersects the y-axis. This occurs when \(x = 0\). To find the y-intercept for a quadratic equation \(y = ax^2 + bx + c\):

• Substitute \(x = 0\) into the equation and simplify.

For our equation \(y = x^2 - 8x + 10\), substituting \(x = 0\) gives us \(y = 10\). Therefore, the y-intercept is (0, 10). This is the point where our parabola crosses the y-axis.

X-Intercept

X-intercepts are the points where the parabola crosses the x-axis. To find the x-intercepts for a quadratic equation \(y = ax^2 + bx + c\), set \(y = 0\) and solve for \(x\). This can be done using the quadratic formula:
\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\] Calculate the discriminant, which determines the number of x-intercepts.

• If the discriminant is positive, there are two x-intercepts.
• If it’s zero, there’s one x-intercept.
• If it’s negative, there are no real x-intercepts.

In our exercise, the discriminant was 24, giving us two real solutions: \(x = 4 + \sqrt{6}\) and \(x = 4 - \sqrt{6}\). This means our parabola intersects the x-axis at these two points.