Problem 12 - Thinking Mathematically

Question

a. Determine if the parabola whose equation is given opens upward or downward. b. Find the vertex. c. Find the \(x\)-intercepts. d. Find the y-intercept. e. Use (a)-(d) to graph the quadratic function. \(f(x)=x^{2}+4 x-5\)

Step-by-Step Solution

Answer

The parabola of the function \(f(x)=x^{2}+4 x-5\) opens upward. The vertex is at (-2,-1), the x-intercepts are at (-5,0) and (1,0), and the y-intercept is at (0,-5).

Step 1: Identify the Opening Direction of the Parabola

The leading coefficient of the quadratic function is 1 which is positive. This means that the parabola opens upward.

Step 2: Identify the Vertex

To find the x-coordinate of the vertex, substitute the coefficients \(a = 1\) and \(b = 4\) into the vertex formula, getting \(h = -\frac{b}{2a} = -\frac{4}{2} = -2\). Substitute \(h = -2\) to the function to get the y-coordinate \(k = f(-2)=(-2)^{2}+4(-2)-5 = -1\). So, the vertex is (-2,-1).

Step 3: Find the x-intercepts

Solve for \(x\) by setting the function equal to 0: \(x^{2}+4 x-5=0\). By applying the quadratic formula \(x = \frac{-b ± \sqrt{b^{2}-4ac}}{2a}\), the solutions gives \(x=-5 and x=1\). These are the x-intercepts of the function.

Step 4: Find the y-intercept

The y-intercept is found by setting \(x = 0\) in the equation: \(f(x) = (0)^{2} + 4(0) - 5 = -5\). So, the y-intercept is -5.

Step 5: Graph the Function

Plot the vertex (-2,-1), the y-intercept (0,-5), and the x-intercepts (-5,0) and (1,0) and draw a parabola that opens upward passing through these points.

Key Concepts

Parabola Opening Direction

The opening direction of a parabola is a fundamental feature to understand when graphing quadratic functions. It tells us if the parabola has a 'smiley face' (opens upward) or a 'frowny face' (opens downward). This characteristic is determined by the sign of the leading coefficient in the quadratic equation of the form f(x) = ax^2 + bx + c. If a is positive, the parabola opens upward, whereas if a is negative, it opens downward.

In our example, with the equation f(x) = x^2 + 4x - 5, the leading coefficient is 1, which is indeed positive. This indicates that our graph will be a 'smiley face' curving upwards from the vertex. Understanding the direction in which a parabola opens is crucial as it affects the position of the vertex and the shape of the graph.

Vertex of a Parabola

The vertex is the highest or lowest point on a parabola, acting as a pivotal element in its graph. It's the point where the parabola changes direction. To find the vertex of a parabola given by the equation f(x) = ax^2 + bx + c, we use the formula h = -b/(2a) to calculate the x-coordinate and then substitute h back into the function to find the y-coordinate. The vertex form of a parabola is then (h, k) where k = f(h).

In our quadratic function f(x) = x^2 + 4x - 5, by applying the vertex formula, we calculated h = -4/(2*1) = -2 and then found k = f(-2) = -1. So, the vertex of this parabola is (-2,-1). Locating the vertex is an essential step towards sketching the graph since it serves as a reference point.

X-intercepts of a Quadratic Function

X-intercepts are the points where the graph of a quadratic function crosses the x-axis, also known as the solutions to the equation. To find the x-intercepts, you set the quadratic equation f(x) = 0 and solve for x. The quadratic formula, x = (-b ± √(b^2-4ac))/(2a), can be used when the equation does not factor easily.

Our function, f(x) = x^2 + 4x - 5, yields the x-intercepts by setting it to zero and solving for x, giving us the points x = -5 and x = 1. These are the points where the parabola meets the x-axis, and they are instrumental in shaping the graph.

Y-intercept of a Quadratic Equation

The y-intercept is the point where the graph intersects the y-axis, which occurs when x is zero. Simply substitute x with 0 in the quadratic equation to find the y-intercept. It's easy to locate and helps to plot the quadratic graph accurately.

For example, the y-intercept of f(x) = x^2 + 4x - 5 is found by calculating f(0) = 0^2 + 4(0) - 5 = -5. Therefore, the point where our parabola crosses the y-axis is (0,-5). This single value can guide students when they start plotting the parabola on a coordinate plane.