Problem 21 - 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. $$ f(x)=-4 x^{2}-16 x+3 $$

Step-by-Step Solution

Answer

Vertex: (2, -45), x-intercepts: (-0.19, 0) and (-3.81, 0), y-intercept: (0, 3); parabola opens downwards.

Step 1: Write Quadratic in Standard Form

The given quadratic function is already in standard form, which is \( f(x) = ax^2 + bx + c \). Here, \( a = -4 \), \( b = -16 \), and \( c = 3 \).

Step 2: Find the Vertex

For a quadratic function \( f(x) = ax^2 + bx + c \), the vertex \((h, k)\) can be found using the formulas \( h = -\frac{b}{2a} \) and \( k = f(h) \).First, find \( h \):\[ h = -\frac{-16}{2(-4)} = 2 \\]Then, find \( k \):\[ k = f(2) = -4(2)^2 - 16(2) + 3 = -16 - 32 + 3 = -45 \\]Thus, the vertex is \((2, -45)\).

Step 3: Calculate the x-intercepts

The x-intercepts occur where \( f(x) = 0 \). Solve the equation \(-4x^2 - 16x + 3 = 0\) using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \).\[ b^2 - 4ac = (-16)^2 - 4(-4)(3) = 256 + 48 = 304 \ x = \frac{-(-16) \pm \sqrt{304}}{2(-4)} = \frac{16 \pm \sqrt{304}}{-8} \\]Calculate the values:- First root: \( x_1 = \frac{16 + \sqrt{304}}{-8} \)- Second root: \( x_2 = \frac{16 - \sqrt{304}}{-8} \)These individual calculations yield approximately \( x_1 \approx -0.19 \) and \( x_2 \approx -3.81 \).

Step 4: Determine the y-intercept

The y-intercept occurs where \( x = 0 \). Calculate \( f(0) \):\[ f(0) = -4(0)^2 - 16(0) + 3 = 3 \\]Thus, the y-intercept is \( (0, 3) \).

Step 5: Sketch the Graph

To sketch the graph:1. Plot the vertex at \((2, -45)\).2. Mark the x-intercepts at approximately \((-0.19, 0)\) and \((-3.81, 0)\).3. Plot the y-intercept at \((0, 3)\).4. Note that the parabola opens downwards since \( a = -4 \), indicating it's a downward facing concave parabola.5. Draw a smooth curve passing through these points.

Key Concepts

Vertex Form

The vertex form of a quadratic function makes it easier to identify the vertex of the parabola quickly. It is given by the expression \( f(x) = a(x - h)^2 + k \), where \((h, k)\) is the vertex of the parabola. For the quadratic function \( f(x) = -4x^2 - 16x + 3 \), the vertex can be calculated using the formulas for finding \( h \) and \( k \).
- First, find the value of \( h \) by using the formula \( h = -\frac{b}{2a} \). In this case, it simplifies to \( h = -\frac{-16}{2(-4)} = 2 \).
- Next, substitute \( h \) back into the function to find \( k \). So, \( k = f(2) = -45 \).
The vertex of this parabola is \((2, -45)\), and knowing this helps us understand where the lowest or highest point of the graph is located. This is particularly useful when graphing parabolas or understanding how the graph shifts.

X-intercepts

X-intercepts are the points where the graph of the quadratic function crosses the x-axis. These occur where \( f(x) = 0 \). Finding these points can help in sketching the graph of the function. Solving the equation \(-4x^2 - 16x + 3 = 0\) gives the x-intercepts.

• The method of solving for x-intercepts often involves the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \).
• For our quadratic: calculate the part under the square root, \( b^2 - 4ac \). This gives \( 304 \) in the expression.
• Solve to find the roots, \( x_1 \approx -0.19 \) and \( x_2 \approx -3.81 \).

These approximate solutions represent the points where the parabola meets the x-axis and are critical in painting a complete picture of the shape of the function's graph.

Y-intercepts

The y-intercept of a function is the point where the graph crosses the y-axis. This can provide a reference point when graphing or analyzing the function. For any quadratic function of the form \( f(x) = ax^2 + bx + c \), the y-intercept occurs where \( x = 0 \). This simplifies the determination of the y-intercept.
For our function \( f(x) = -4x^2 - 16x + 3 \):
- Substitute \( x = 0 \) into the function to find \( f(0) = 3 \).
Thus, the y-intercept is \((0, 3)\). This point is essential as it shows where the parabola begins its journey on the y-axis, providing a starting point while sketching the graph.

Graphing Parabolas

Graphing parabolas involves using key points such as the vertex, x-intercepts, and y-intercept to draw a smooth curve that represents the function. Understanding the direction in which the parabola opens is crucial as well.

• First, plot the vertex \((2, -45)\), which serves as the "tip" of the parabola.
• Next, mark the x-intercepts approximately at \((-0.19, 0)\) and \((-3.81, 0)\).
• Find and plot the y-intercept at \((0, 3)\).

Quintessential to note is that if \( a < 0 \), the parabola will open downwards. Here, \( a = -4 \), so indeed, it opens downwards. Draw a smooth curve through all these crucial points to complete the parabola graph. This visualization represents the function's behavior and helps in predicting function values between known points.