Problem 93 - Precalculus : Building Concepts and Connections

Question

Suppose that the vertex and an \(x\) -interceptl of the parabola associated with a certain quadratic function are given by (-1,2) and \((4,0),\) respectively. (a) Find the other \(x\) -intercept. (b) Find the equation of the parabola. (c) Check your answer by graphing the function.

Step-by-Step Solution

Answer

The other x-intercept is (-6,0). The equation of the parabola is \(y=-0.2(x+1)^2+2\)

Step 1: Find the other x-intercept

Due to the symmetry property of parabolas, the other x-intercept will be the same distance from the vertex but in the opposite direction. The given x-intercept is 4, and the vertex is -1. The distance between them is \(4-(-1) = 5\). The other x-intercept will be \( -1 - 5 = -6\), implying the other point is (-6,0).

Step 2: Find the equation of the parabola

The equation of a parabolic function is given by \(y=a(x-h)^2+k\). The vertex is given by (h,k), which is (-1, 2) in this case. So far we have \(y=a(x+1)^2+2\). To find 'a', substitute the coordinates of a known point into the equation, the given x-intercept (4,0) for instance. 0=a(4+1)^2+2 gives a=-0.2.

Step 3: Write the equation

Finalize the equation of the parabola using the a, h and k values obtained: \(y=-0.2(x+1)^2+2\)

Step 4: Graph the function

To verify your answer, you should graph the function \(y=-0.2(x+1)^2+2\). The parabola opens downward (since 'a' is negative), the vertex is at (-1,2), and the x-intercepts are at (4,0) and (-6,0), which confirms the accuracy of your answer. Manual graphing is not feasible in this format but software or graphing calculators can be used to check the function.

Key Concepts

Parabola Vertex

The parabola vertex is a crucial point to understand on a graph. It's the point where the parabola changes direction and is considered the minimum or maximum point of the curve, depending on whether the parabola opens upwards or downwards. In our exercise, the vertex is given as (-1, 2). This means the parabola reaches its peak or trough at this point. The vertex is used in the vertex form of a quadratic equation, which is presented as \( y = a(x-h)^2 + k \), where \((h, k)\) is the vertex. Understanding the vertex helps tremendously when graphing a parabola because it provides a clear center point around which the rest of the curve is shaped. The vertex tells us not only where the turning point is but also helps in quickly deducing the direction in which the parabola opens, based on the value of \(a\). If \(a\) is negative, the parabola opens downward, making the vertex the highest point, which applies to our solution.

X-Intercepts

X-intercepts are points where the graph of a function crosses the x-axis, which means the y-value is zero at these points. For a parabola, there can be zero, one, or two x-intercepts. In our problem, one x-intercept is given as (4, 0). To find the other x-intercept, we utilize the symmetry of the parabola. Since the x-intercepts are equidistant from the vertex along the x-axis, we can calculate the second x-intercept. The vertex is at \(-1\), and the given x-intercept is \(4\). The distance between \(-1\) and \(4\) is \(5\), so moving \(5\) units in the opposite direction from the vertex gives another x-intercept at \(-6\). This symmetry is a vital property, allowing us to determine the second x-intercept easily and ensuring we have all key points for graphing the function accurately.

Equation of a Parabola

The equation of a parabola in vertex form is \( y = a(x-h)^2 + k \), where \((h, k)\) are the coordinates of the vertex. In our exercise, we have the vertex \((-1, 2)\), which integrates into our equation as \( y = a(x+1)^2 + 2 \). The next step is determining \(a\). Using an x-intercept, such as \((4, 0)\), to substitute into the equation helps find this value of \(a\). By setting \(y\) to zero (since it's an intercept), we solve \(0 = a(4+1)^2 + 2\), yielding \(a = -0.2\). Thus, the parabolic equation for our function becomes \( y = -0.2(x+1)^2 + 2 \). This equation not only describes the shape of the parabola but also encapsulates vital properties such as its direction (downward, since \(a\) is negative) and position on the coordinate plane. It's crucial to master setting up and adjusting this equation as it is fundamental for analyzing quadratic functions.