Problem 17 - Introductory Algebra for College Students

Question

Find the \(y\) -intercept for the parabola whose equation is given. $$y=x^{2}+6 x$$

Step-by-Step Solution

Answer

The y-intercept of the parabola \(y = x^2 + 6x\) is \(y = 0\).

Step 1: Substitute zero for x

We substitute \(x = 0\) into the equation \(y = x^{2} + 6x\) to find the y-intercept. Hence the equation becomes \(y = 0^{2} + 6*0\)

Step 2: Simplify the equation

Simplifying the equation, we get \(y = 0\). Thus, the y intercept for the equation is 0.

Key Concepts

Parabola Equation

When we talk about a parabola equation, we're often referring to a quadratic equation in its standard form, which is \( y = ax^2 + bx + c \).

This equation represents a parabola on a coordinate plane, where \(a\), \(b\), and \(c\) are constants. It's important to note that if \(a > 0\), the parabola opens upward, and if \(a < 0\), the parabola opens downward. The constant \(c\) represents the y-intercept, which is the point where the parabola crosses the y-axis. Therefore, finding the y-intercept of a parabola is as simple as looking at the value of \(c\) in the equation.

To clarify with an example, for the equation \(y = x^2 + 6x\), the y-intercept is found by setting the value of \(x\) to zero. This will eliminate the \(x\)-terms and leave us with \(c\) as the answer: \(y = 0^2 + 6\cdot0 = 0\). Thus, we've found the y-intercept without needing to graph the equation first.

Graphing Parabolas

When it comes to graphing parabolas, understanding the structure of the quadratic equation is key. The general form of the parabola equation is incredibly helpful, as it gives us points to plot on the graph, such as the vertex and the y-intercept.

The vertex of a parabola is the highest or lowest point on the graph and occurs at \(x = -\frac{b}{2a}\) when the equation is in standard form. After finding the vertex, you can plot additional points by choosing x-values and calculating their corresponding y-values. Symmetry is a factor as well: for every point on one side of the vertex, there is a mirrored point on the other side.

For our example equation, \(y = x^2 + 6x\), we already know the y-intercept is at (0,0). To graph the parabola, one can complete the square or find the vertex through the formula given above, then plot points on each side, and draw a smooth curve to complete the graph. Remember, always plot at least one point to the left and to the right of the vertex to ensure accuracy in the parabola's shape.

Solving Quadratic Equations

Solving quadratic equations is an essential skill for understanding parabolas. These equations can be solved through several methods, including factoring, using the quadratic formula, completing the square, or graphically.

Factoring is most useful when the equation can be easily expressed as a product of binomials. If the equation does not factor easily, the quadratic formula, \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), can always be used to find the x-intercepts of the parabola. Completing the square involves rewriting the equation so that the left side is a perfect square trinomial, allowing for easy solving of \(x\). Graphically, one can solve for \(x\) by finding the points where the parabola intersects the x-axis.

In our example where \(y = x^2 + 6x\), factoring might be the first method to try, though it is clear by inspection that this particular equation does not factor nicely, suggesting that completing the square or using the quadratic formula might be better approaches to find the values of \(x\) when \(y\) equals zero (the x-intercepts).