Question

Sketch the function \(f(x)=x^2+6 x+8\), and list the intercepts, if any, with the \(x\) and \(f(x)\) axes.

Short answer

The intercepts are (0, 8), (-2, 0), and (-4, 0), and the vertex is (-3, 1).

Step-by-step solution

- Find the Intercepts

To find the intercepts with the axes, set up and solve the necessary equations. For the y-intercept, set the value of x to 0 in the function: \[ f(0) = 0^2 + 6 \times 0 + 8 = 8 \]. So, the y-intercept is at (0, 8). For the x-intercept, set the function equal to 0: \[ x^2 + 6x + 8 = 0 \]. Solve the quadratic equation to find the x-values where the function equals zero.

- Solve the Quadratic Equation

Use factoring or the quadratic formula to solve the equation: Factorizing the quadratic, we get: \[ (x + 2)(x + 4) = 0 \]. Therefore, the solutions are: \[ x = -2 \] and \[ x = -4 \]. So, the x-intercepts are (-2, 0) and (-4, 0).

- Identify the Vertex

The vertex form of a quadratic equation \[ ax^2 + bx + c \] can be found using \[ x = -\frac{b}{2a} \]. Here, \[ a = 1 \] and \[ b = 6 \]. Thus, the vertex is \[ x = -\frac{6}{2 \times 1} = -3 \]. Substitute \[ x = -3 \] back into the function to find the y-coordinate: \[f(-3) = (-3)^2 + 6(-3) + 8 = 1 \]. So, the vertex is (-3, 1).

- Sketch the Function

Plot the points found: the y-intercept (0, 8), the x-intercepts (-2, 0) and (-4, 0), and the vertex (-3, 1). These points guide the sketch. Since the coefficient of \(x^2\) is positive, the parabola opens upwards. Draw the parabola through these points.

Key concepts

Quadratic Equation Intercepts

When graphing quadratic functions, finding the intercepts is a crucial step. Intercepts are points where the graph crosses the axes.
To find the y-intercept, set the value of x to 0. This tells us where the graph crosses the y-axis.
For our function, we set x to 0: \[ f(0) = 0^2 + 6 \times 0 + 8 = 8 \].
So, the y-intercept is (0, 8).
For the x-intercepts, we solve for x when the function equals 0: \[x^2 + 6x + 8 = 0 \].
This equation can be solved by factoring: \[ (x + 2)(x + 4) = 0 \].
Therefore, we find that x = -2 and x = -4.
The x-intercepts are the points (-2, 0) and (-4, 0). Finding intercepts helps us understand key points where the graph intersects the axes, providing a foundation for the graph's shape.

Vertex of a Parabola

The vertex of a parabola is an essential point that defines its maximum or minimum. For any quadratic function in the form \[ax^2 + bx + c \], the vertex can be found using the formula for x: \[ x = -\frac{b}{2a} \].
In our function, \[f(x) = x^2 + 6x + 8 \], we have \[ a = 1 \] and \[ b = 6 \].
Plugging these values into our formula: \[ x = -\frac{6}{2 \times 1} = -3 \].
To find the corresponding y-coordinate, substitute \[ x = -3 \] back into the function: \[ f(-3) = (-3)^2 + 6(-3) + 8 = 1 \].
So, the vertex is at (-3, 1).
The vertex shows the highest or lowest point on the graph, and for parabolas opening upwards, it is the lowest point. The vertex helps in sketching the curve and understanding the parabola's orientation.

Sketching Parabolas

To sketch a parabola, gather key points: the y-intercept, x-intercepts, and vertex.
In our example, we have the y-intercept at (0, 8), x-intercepts at (-2, 0) and (-4, 0), and the vertex at (-3, 1).
These points guide the shape of the graph.
Start by plotting these points on a coordinate plane.
Since the coefficient of \[x^2\] is positive in our function, \[f(x) = x^2 + 6x + 8 \], the parabola opens upwards.
Draw a smooth curve through the plotted points to illustrate the parabola.
Remember, the vertex represents the lowest point, and the graph is symmetric around the line \[ x = -3 \], which is the vertex's x-coordinate.
By connecting these points, you create an accurate sketch of the quadratic function. The curve's direction and shape become clear using these crucial points as reference.