Problem 21 - Algebra and Trigonometry

Question

Use the vertex and intercepts to sketch the graph of each quadratic function. Give the equation of the parabola's axis of symmetry. Use the graph to determine the function's domain and range. $$ y-1=(x-3)^{2} $$

Step-by-Step Solution

Answer

The vertex of the parabola is (3,1), it has no x-intercepts. The axis of symmetry is the line x=3. The domain of the function is all real numbers, and the range of the function is [1,+\infty). The graph opens upwards with the vertex at the point (3,1), indicates that parabola has a minimum value at y=1.

Step 1: Find the vertex of the parabola

The equation of the parabola is in the form \( y = a(x-h)^2 + k \), where (h,k) is the vertex. From the equation \( y - 1 = (x-3)^2 \), it can easily be seen that the vertex (h,k) is (3,1).

Step 2: Calculate the x-intercepts

The x-intercepts are determined by setting y equal to zero and solving for x. Let's do this: \[ 0 - 1 = (x-3)^2 \] \[ -1 = (x-3)^2 \] Since the square of a real number is non-negative, this equation has no solution. Therefore, the parabola does not have an x-intercept.

Step 3: Determine the axis of symmetry

The axis of symmetry of a parabolic graph is a vertical line through the vertex. Therefore, in this case, it is the line x=3.

Step 4: Determine the function's domain and range

The domain of the function is all real numbers. The range of a parabolic function of the form \( y = a(x-h)^2 + k \) is \[ k, + \infty ) \] if a is positive. Since, in this case, the vertex is located above the x-axis, the range is \( [1, +\infty)\).

Step 5: Sketch the graph

Start with the vertex point (3,1), sketch the axis of symmetry and indicate that the parabola opens up. Since there are no x-intercepts, the parabola only touches the y-axis at the vertex point.

Key Concepts

Vertex

In a quadratic function, the vertex is a key point that represents either the minimum or maximum value of the function, depending on the direction the parabola opens. In equations of the form \( y = a(x-h)^2 + k \), the vertex is the point \((h, k)\).

For the quadratic function given by \( y - 1 = (x-3)^2 \), we can identify the vertex by comparing it to the standard vertex form of a quadratic function. Here, our equation tells us that the vertex is \((3, 1)\).

The vertex offers valuable information. It acts as an anchor point in determining the shape and position of the parabola on a graph.

Axis of Symmetry

The axis of symmetry of a parabola is an imaginary line that vertically slices the parabola into two mirror-image halves.

This line always passes through the vertex of the parabola. Given a vertex at \((h, k)\), the axis of symmetry is defined by the equation \(x = h\).

For our quadratic function \(y - 1 = (x-3)^2\), where the vertex is \((3, 1)\), the axis of symmetry is the vertical line \(x = 3\). This line helps in graphing the parabola and understanding its symmetric nature, making it easier to plot points on either side of the axis.

Domain and Range

The domain of a quadratic function is a set of all possible x-values that can be input into the function, which, for any quadratic function, is all real numbers \((-\infty, +\infty)\). This means we can choose any value for \(x\) and find a corresponding \(y\).

The range of a quadratic function refers to all the possible y-values the function can output. For parabolas, this depends on the direction they open. When the parabola opens upwards (as is the case in our function since the term \((x-3)^2\) indicates there's no negative factor), the range starts at the vertex's y-value and extends to infinity.

Therefore, for the parabola defined by \(y - 1 = (x-3)^2\), the range is all y-values from the vertex upward - that is, \([1, +\infty)\).

Parabola

A parabola is the graph of a quadratic function. It is always U-shaped and can open either upwards or downwards.

The direction in which the parabola opens is determined by the coefficient \(a\) in the standard equation \(y = a(x-h)^2 + k\). If \(a > 0\), the parabola opens upwards, and if \(a < 0\), it opens downwards.

In our equation \(y - 1 = (x-3)^2\), the absence of a leading negative sign in \((x-3)^2\) implies that \(a\) is positive, suggesting that the parabola opens upwards.

A well-drawn parabola should clearly depict its vertex, and having accurately plotted the vertex, the axis of symmetry helps maintain its shape as a symmetrical curve on either side of this axis.