Problem 9 - Algebra and Trigonometry

Question

A quadratic function is given. (a) Express the quadratic function in standard form. (b) Find its vertex and its x- and y-intercept(s). (c) Sketch its graph. $$ f(x)=x^{2}-6 x $$

Step-by-Step Solution

Answer

The standard form is \( f(x) = (x-3)^2 - 9 \); vertex is \((3, -9)\); x-intercepts are \(0\) and \(6\); y-intercept is \((0,0)\). Sketch based on these points.

Step 1: Express the Function in Standard Form

The general form of a quadratic function is given by \( f(x) = ax^2 + bx + c \). Here, we have \( f(x) = x^2 - 6x \). To complete the square and express this in standard form \( f(x) = a(x-h)^2 + k \), rewrite the function. First, factor out the leading coefficient if necessary (which is 1 in this case, so we skip factoring). Then, complete the square:1. Take half of the coefficient of \( x \), which is \(-6\), divide by 2 getting \(-3\). Square it to get \( 9 \).2. Add and subtract this number inside the function: \( f(x) = x^2 - 6x + 9 - 9 \).3. Rewrite the expression as a perfect square: \( f(x) = (x-3)^2 - 9 \).

Step 2: Identify the Vertex

The vertex form of our quadratic function is \( f(x) = (x-3)^2 - 9 \). From this, we can identify the vertex \((h, k)\), which is \((3, -9)\) because the vertex form is \( a(x-h)^2 + k \). Thus, the vertex is at the point \((3, -9)\).

Step 3: Find the x-intercepts

To find the x-intercepts, set \( f(x) = 0 \) and solve for \( x \):\((x-3)^2 - 9 = 0 \)1. Add 9 to both sides: \( (x-3)^2 = 9 \).2. Take the square root of both sides: \( x-3 = \pm 3 \).3. Solve for \( x \): \( x = 3 + 3 = 6 \) and \( x = 3 - 3 = 0 \).Thus, the x-intercepts are at \( x = 6 \) and \( x = 0 \).

Step 4: Find the y-intercept

The y-intercept occurs when \( x = 0 \). Substitute \( x = 0 \) into the original function:\( f(0) = (0)^2 - 6(0) = 0 \).Thus, the y-intercept is at the point \((0, 0)\).

Step 5: Sketch the Graph

To sketch the graph of the function, plot the vertex \((3, -9)\) and the intercepts \((0, 0)\) and \((6, 0)\). The parabola opens upwards because the coefficient of \( x^2 \) is positive (1). Draw a symmetric curve passing through these points, with axis of symmetry at \( x = 3 \). The points plotted and symmetry axis will help accurately sketch the curve.

Key Concepts

Vertex Form

Quadratic functions can be challenging to interpret in their standard form. This is where the vertex form comes in handy. The vertex form of a quadratic function is written as \[ f(x) = a(x-h)^2 + k \]where

• \(a\) determines the direction and the width of the parabola
• \(h\) and \(k\) represent the vertex coordinates \((h, k)\)

To convert the given quadratic function into vertex form, you use the method of completing the square. For example, with the function \[ f(x) = x^2 - 6x \]the completed square form becomes \[ f(x) = (x-3)^2 - 9 \].This reveals that the vertex of the parabola is at the point \((3, -9)\). Knowing the vertex helps us understand the location and shape of the parabola graph.

X-Intercepts

Finding x-intercepts involves setting the function equal to zero. For the quadratic function \[ f(x) = (x-3)^2 - 9 \],you solve \[ (x-3)^2 - 9 = 0 \].Adding 9 to both sides, we get \[ (x-3)^2 = 9 \].By taking the square root, you simplify to \[ x-3 = \pm 3 \].This leads to two values of \( x \):

• \( x = 6 \)
• \( x = 0 \)

Thus, the parabola crosses the x-axis at \( x = 6 \) and \( x = 0 \). These intercepts are crucial because they show where the graph intersects the x-axis. This is helpful when sketching the overall shape of the parabola.

Y-Intercept

The y-intercept of a function is the point where the graph crosses the y-axis. For any function, you find this by setting \( x = 0 \) in the equation. Applying this to our function, we substitute \( x = 0 \) into \[ f(x) =x^2 - 6x \].Simplifying, we calculate \[ f(0) = 0^2 - 6(0) = 0 \].This shows us that the parabola passes through the origin, at the point \((0, 0)\).The y-intercept serves as a starting point to understand the function's behavior. It is particularly useful in sketching the graph because it provides a fixed point through which the curve must pass.

Graph Sketching

Sketching the graph of a quadratic function involves gathering all the information about the function, such as the vertex, x-intercepts, and y-intercept. For the function \[ f(x) = (x-3)^2 - 9 \],the vertex is \((3, -9)\), the x-intercepts are at \( x = 0 \) and \( x = 6 \), and the y-intercept is \((0, 0)\).When sketching the graph, plot these points first:

• The vertex \((3, -9)\) shows the lowest point on the graph since the parabola opens upwards.
• The x-intercepts \((0, 0)\) and \((6, 0)\) indicate where the graph crosses the x-axis.
• The y-intercept \((0, 0)\) is the same as one of the x-intercepts in this case.

Making use of these points, draw a symmetric parabola that opens upwards. Remember, the axis of symmetry for this parabola is \(x = 3\). This axis helps ensure that your sketch is properly centered, with reflections across this central line ensuring an accurate representation of the function's graph.