Problem 44 - Beginning and Intermediate Algebra

Question

Graph each equation using the vertex formula. Find the \(x\) - and \(y\) -intercepts. $$x=-y^{2}+4 y$$

Step-by-Step Solution

Answer

The equation \(x = -y^2 + 4y\) in vertex form is \(x = -(y - 2)^2 + 4\). The vertex of the parabola is at point (4,2). The y-intercepts are at the points (0,0) and (0,4), and the x-intercept is at the point (0,0). The parabola opens to the left and passes through the intercept points.

Step 1: Express equation in the vertex form #

Rewrite the given equation in the form \(x = a(y-h)^2 + k\) where (h,k) is the vertex of the parabola. Now, let's complete the square in terms of y : \(x = -(y^2 - 4y) = -(y^2 - 4y + (4/2)^2) + 4(4/2)\) So the equation in vertex form is: \(x = -(y - 2)^2 + 4\) The vertex of this parabola is (h,k) = (4,2).

Step 2: Find y-intercept #

To find the y-intercept, we set x to 0 and solve for y. \(0 = -(y - 2)^2 + 4\) Now, rearrange and solve for y: \((y - 2)^2 = 4\) \(y- 2 = \pm\sqrt{4}\) \(y = 2 \pm \sqrt{4}\) So there will be two y-intercepts at the points: (0,0) and (0,4)

Step 3: Find x-intercept #

To find the x-intercept, set y to 0 and solve for x. \(x = -(0 - 2)^2 + 4\) \( x = -4 + 4\) The x-intercept is at the point: (0,0)

Step 4: Graph the equation #

Now, graph the equation with the vertex (4,2) and the x- and y-intercepts found. The parabola will have a vertex point at (4,2) and it will open to the left since the coefficient of the y^2 term is negative. The graph will pass through the x-intercept (0,0) and y-intercepts (0,0) and (0,4). This exercise shows how to graph a quadratic equation using the vertex formula and finding both the x-intercept and the y-intercept.

Key Concepts

Vertex Formula

The vertex formula is a powerful tool when dealing with quadratic equations. It allows you to rewrite these equations in a way that makes it easier to identify the vertex of a parabola. In general, the vertex form of a quadratic equation is given by:\[ x = a(y-h)^2 + k \]Here, \(h,k\) represents the vertex of the parabola. The parameter \(a\) affects the width and direction of the parabola. For our equation, rewriting in vertex form is achieved by completing the square, resulting in:\[ x = -(y - 2)^2 + 4 \]This tells us that the vertex \(h,k\) is located at the point (4,2).Understanding the vertex's location helps in sketching the graph accurately. Remember, finding the vertex is crucial because it represents the maximum or minimum point of the parabola, depending on its orientation.

Parabola Graphing

Graphing a parabola when given its vertex form can be straightforward. First, determine the vertex, which in this case is at (4,2). This is the starting point of your graph. The equation also tells us that because the coefficient in front of \(y^2\) is negative, the parabola opens to the left. To graph a parabola effectively:

• Start with plotting the vertex.
• Determine the direction the parabola opens by looking at the sign of the \(a\) value (negative opens left for horizontal parabolas).
• Use any known intercepts to assist in plotting points on the graph.
• Sketch the curve, ensuring it passes through these critical points.

With this process, you'll have a clear, accurate graph of the quadratic equation in question.

X-Intercepts

Finding the x-intercepts involves setting the value of y to zero in the quadratic equation. This is because x-intercepts are points where the graph touches or crosses the x-axis, meaning the y-value at these points is zero. For the given exercise, when we set y = 0, we found:\[ x = -(0 - 2)^2 + 4 \]\[ x = 0 \]Thus, the x-intercept is at the point (0,0). Having x-intercepts helps in verifying the accuracy of the plotted parabola on the graph. Remember, identifying the x-intercept is crucial for understanding where the graph interacts with the x-axis, which is a key component of graph interpretation.

Y-Intercepts

To find the y-intercepts of a parabola, we set the value of x to zero in the equation. This gives us the points where the graph crosses the y-axis, as the x-value is zero at these intersections. In the exercise:\[ 0 = -(y - 2)^2 + 4 \]We solved this to obtain:\[ (y - 2)^2 = 4 \]\[ y - 2 = \pm 2 \]\[ y = 2 \pm 2 \]So, the y-intercepts are at the points (0,0) and (0,4). Understanding these intercepts further assists in the graphical representation, giving clear indications of where the parabola intersects the y-axis. In parabola graphing, such intercepts are critical for confirming the shape and position of the curve.