Problem 32 - College Algebra

Question

For the following exercises, use the vertex \((h, k)\) and a point on the graph \((x, y)\) to find the general form of the equation of the quadratic function. $$ (h, k)=(0,1),(x, y)=(1,0) $$

Step-by-Step Solution

Answer

The equation of the quadratic function is \( f(x) = -x^2 + 1 \).

Step 1: Identify the Vertex Form

The vertex form of a quadratic equation is given by \( f(x) = a(x-h)^2 + k \) where \((h, k)\) is the vertex of the parabola. In our case, \( h = 0 \) and \( k = 1 \), so the equation becomes \( f(x) = a(x-0)^2 + 1 = ax^2 + 1 \).

Step 2: Substitute the Point into the Equation

Using the point \((x, y) = (1, 0)\), substitute \( x = 1 \) and \( y = 0 \) into the equation. This gives us:\[ 0 = a(1)^2 + 1 \]

Step 3: Solve for \(a\)

Solve the equation from the previous step to find the value of \(a\). This becomes:\[ 0 = a(1) + 1 \]\[ 0 = a + 1 \]Subtract 1 from both sides:\[ a = -1 \]

Step 4: Write the General Form

Now that we know \(a = -1 \), substitute this back into the vertex form equation. The equation is now:\[ f(x) = -x^2 + 1 \]This is the general form of the quadratic function given the vertex and point.

Key Concepts

Vertex Form

The vertex form of a quadratic function is an essential concept in understanding parabolas. It's particularly helpful when you know the vertex of the parabola. The vertex form of a quadratic equation is expressed as: \[ f(x) = a(x-h)^2 + k \]Here:

• \((h, k)\) is the vertex of the parabola, representing the point where the parabola either reaches a maximum or minimum value.
• \(a\) determines the direction and width of the parabola. If \(a\) is positive, the parabola opens upwards, and if it's negative, it opens downwards.

In the exercise, the vertex is given as \((h, k) = (0, 1)\). Substituting these values into the vertex form gives: \[ f(x) = a(x-0)^2 + 1 \], which simplifies to \[ ax^2 + 1 \]. This simplified form is helpful for finding the precise equation of the parabola when you also know another point on the graph.

Solving for Coefficients

After establishing the vertex form of a quadratic equation, the next step is to solve for the coefficient \(a\). This can be achieved by using an additional known point on the parabola. In our example, we're given the point \((x, y) = (1, 0)\). By substituting \(x = 1\) and \(y = 0\) into the equation \( f(x) = ax^2 + 1 \), you have:\[0 = a(1)^2 + 1\]This simplifies to \( 0 = a + 1 \). Solving this equation involves the following steps:

• Isolate \(a\) by subtracting 1 from both sides: \( a = -1 \).

Now, you understand that \(a = -1\), this tells us that our parabola opens downwards.

Quadratic Equation Graphing

Once you know the quadratic equation, graphing becomes more intuitive. In our example, we've established the equation as \( f(x) = -x^2 + 1 \). To graph a quadratic equation, follow these steps:

• Identify the vertex of the parabola from the vertex form. Here it is \((0,1)\).
• Recognize the direction the parabola opens, which is downwards in this case since \(a = -1\).
• Find additional points by substituting other values of \(x\) into \( f(x) \) to get corresponding \(y\) values.

With the vertex and main axis identified, plot these points on a graph axis. Through these steps, you can visualize how the quadratic equation forms a parabola, showcasing its symmetry and key features, such as intercepts and the vertex.