Problem 17 - Algebra and Trigonometry

Question

\(5-18=\) 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. $$ f(x)=-4 x^{2}-16 x+3 $$

Step-by-Step Solution

Answer

Vertex: (2, -45); y-intercept: (0, 3); Find x-intercepts using the quadratic formula. Graph the function.

Step 1: Express the Quadratic Function in Standard Form

A quadratic function is typically given as \(f(x) = ax^2 + bx + c\), which is its standard form. The equation \(-4x^2 - 16x + 3\) is already in this form. Identify \(a = -4\), \(b = -16\), and \(c = 3\).

Step 2: Find the Vertex

The vertex formula for a quadratic function \(f(x) = ax^2 + bx + c\) is \(x = -\frac{b}{2a}\). Substituting \(a = -4\) and \(b = -16\), we get \(x = -\frac{-16}{2(-4)} = 2\). To find the \(y\)-coordinate, substitute \(x = 2\) back into the function: \(f(2) = -4(2)^2 - 16(2) + 3 = -4(4) - 32 + 3 = -16 - 32 + 3 = -45\). The vertex is \((2, -45)\).

Step 3: Find the x-intercepts

Set \(f(x) = 0\) to find the \(x\)-intercepts: \[-4x^2 - 16x + 3 = 0\]Use the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\). For this function, \(a = -4\), \(b = -16\), and \(c = 3\). Calculate the discriminant:\(b^2 - 4ac = (-16)^2 - 4(-4)(3) = 256 + 48 = 304\). So, \(x = \frac{16 \pm \sqrt{304}}{-8}\), which gives solutions for \(x\).

Step 4: Find the y-intercept

Set \(x = 0\) in the function to find the \(y\)-intercept: \(f(0) = -4(0)^2 - 16(0) + 3 = 3\). So, the \(y\)-intercept is \((0, 3)\).

Step 5: Sketch the Graph

Given the vertex \((2, -45)\), the \(x\)-intercepts found in Step 3, and the \(y\)-intercept \((0, 3)\), sketch the graph. The parabola opens downwards (since \(a = -4 < 0\)) and is symmetric about the vertex. Use these intercepts and the vertex to plot the graph.

Key Concepts

Vertex Form

The vertex form of a quadratic function is incredibly useful for graphing and understanding the basic shape and location of a parabola. It is given by the expression \(f(x) = a(x-h)^2 + k\), where

• \(a\) represents the same coefficient as in the standard form, determining the parabola's width and direction.
• \((h, k)\) represent the vertex of the parabola, where \(h\) is the x-coordinate and \(k\) is the y-coordinate.

The vertex form allows you to easily see the vertex's position on a graph, making it straightforward to sketch the parabola quickly. For example, if you have a function like \(f(x)=-4(x-2)^2 + 5\), the vertex would be at point \((2,5)\). This form is especially helpful if you need to graph the parabola or understand its minimum or maximum point (the vertex). Always make sure to convert between standard and vertex forms if needed to get a clearer picture of the quadratic's graph.

X-Intercepts

The x-intercepts of a quadratic function are points where the graph crosses the x-axis. To find these, you set \(f(x) = 0\) and solve the resulting quadratic equation. This can be done using factoring, completing the square, or the quadratic formula:

• Quadratic Formula: \[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]

Using our example function, \(-4x^2 - 16x + 3 = 0\), plug in the values \(a = -4\), \(b = -16\), and \(c = 3\) to find the discriminant. After calculating, the x-intercepts can be found by solving:

• \(x = \frac{16 \pm \sqrt{304}}{-8}\)

The two solutions for \(x\) give the exact points where the parabola intersects the x-axis. These intercepts are important for graphing as they outline the range where the function takes positive and negative values across the domain.

Y-Intercepts

Finding the y-intercept of a quadratic function is simpler than finding the x-intercepts. It's the point where the graph crosses the y-axis, which occurs when \(x = 0\). Simply substitute \(x = 0\) into the function to find the y-intercept. For the equation\(-4x^2 - 16x + 3\), set \(x = 0\), getting:

• \(f(0) = -4(0)^2 - 16(0) + 3 = 3\)

Therefore, the y-intercept is \((0, 3)\). On a graph, this intercept serves as a crucial reference point for the location of the parabola. Since it is usually easier to calculate than x-intercepts, it gives an immediate impression of how the function behaves at the origin.

Graphing Parabolas

Graphing parabolas can initially seem daunting, but with a few key elements, it becomes much easier. You start by identifying the function's vertex and intercepts as they form the core reference points for sketching the parabola. For example, given the vertex \((2, -45)\) and intercepts at \((x_1, 0)\) and \((0, 3)\), you can draft the graph.Begin by plotting the vertex, which indicates the parabola's peak or trough. Next, sketch the x and y-intercepts on the graph, ensuring that the parabola traverses these points. With these fixed points:

• The vertex facilitates symmetry.
• The x and y-intercepts determine width and stretch.

Because the coefficient \(a = -4\) is negative, the parabola opens downwards, indicating a maximum vertex. Use the symmetry of the vertex to reflect points across the axis of symmetry. Naturally, the function's domain is all real numbers; however, the range is limited depending on whether the parabola opens upward or downward.