Problem 47 - Algebra for College Students

Question

Find the vertex and intercepts for each quadratic function. Sketch the graph, and state the domain and range. $$y=-x^{2}-4 x-3$$

Step-by-Step Solution

Answer

Vertex: (2, -15); y-intercept: (0, -3); x-intercepts: (-3, 0) and (-1, 0); Domain: (-∞, ∞); Range: (-∞, -15]

Step 1 - Find the Vertex

The vertex form of a quadratic equation is given by: \[y = a(x-h)^2 + k\] where \((h, k)\) is the vertex. To find the vertex of the quadratic function \(y = -x^2 - 4x - 3\), use the formula for \(h\), which is \(-\frac{b}{2a}\). Here, \(a = -1\) and \(b = -4\).\[h = -\frac{-4}{2(-1)} = 2\]To find \(k\), substitute \(h\) back into the equation:\[k = -h^2 - 4h - 3 = -2^2 - 4(2) - 3 = -4 - 8 - 3 = -15\]Thus, the vertex is \((2, -15)\).

Step 2 - Find the y-intercept

The y-intercept occurs when \(x = 0\). Substitute \(x = 0\) into the equation\(y = -x^2 - 4x - 3\): \[y = -(0)^2 - 4(0) - 3 = -3\]Therefore, the y-intercept is \( (0, -3) \).

Step 3 - Find the x-intercepts

The x-intercepts occur when \(y = 0\). Set \(y\) to 0 in the equation and solve for \(x\):\[0 = -x^2 - 4x - 3\]This is a quadratic equation in standard form. Use the quadratic formula \[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]Here, \(a = -1\), \(b = -4\), and \(c = -3\):\[x = \frac{-(-4) \pm \sqrt{(-4)^2 - 4(-1)(-3)}}{2(-1)}\]\[x = \frac{4 \pm \sqrt{16 - 12}}{-2}\]\[x = \frac{4 \pm 2}{-2}\]So the x-intercepts are:\[x = \frac{6}{-2} = -3 \ x = \frac{2}{-2} = -1\]Therefore, the x-intercepts are \((-3, 0)\) and \((-1, 0)\).

Step 4 - State the Domain and Range

The domain of any quadratic function is all real numbers. Thus:\[Domain: (-\infty, \infty)\]Since this is a downward-facing parabola (the coefficient of \(x^2\) is negative), the range is limited to all values less than or equal to the y-coordinate of the vertex (which is -15). Thus:\[Range: (-\infty, -15]\]

Step 5 - Sketch the Graph

Plot the vertex \((2, -15)\), the y-intercept \((0, -3)\), and the x-intercepts \((-3, 0)\) and \((-1, 0)\). Draw a parabola opening downwards that passes through these points.\[Graph: Draw this on a coordinate plane\]

Key Concepts

Vertex Form

In quadratic functions, the vertex form is an alternative to the standard form and is very useful for quickly identifying the vertex. The vertex form of a quadratic equation is given as: \( y = a(x-h)^2 + k \), where \((h, k)\) represents the vertex of the parabola.
For example, let's take the quadratic function \( y = -x^2 - 4x - 3 \). To convert this to vertex form, we complete the square.
First, remember the vertex formula which utilizes \( h = -\frac{b}{2a} \). Here we get:

• \( h = -\frac{-4}{2(-1)} = 2 \)
• Then, substituting \( h \) back into our equation, \( k = -h^2 - 4h - 3 \), we find:
• \( k = -(2)^2 - 4(2) - 3 = -4 - 8 - 3 = -15 \)

Hence, the vertex form is \( y = -(x-2)^2 - 15 \), placing our vertex at \((2, -15)\).

Y-Intercept

The y-intercept of a quadratic function is where the graph crosses the y-axis. At this point, \(x\) is always 0. To find the y-intercept, substitute \(x = 0\) into the quadratic equation.
For instance, with the equation \(y = -x^2 - 4x - 3\), substituting \(x = 0\) gives us:

• \(y = -(0)^2 - 4(0) - 3 = -3\)

Therefore, the y-intercept is \( (0, -3) \). This means that the graph will cross the y-axis at the point where y is -3.

X-Intercept

The x-intercepts of a quadratic function are points where the graph intersects the x-axis. At these points, \(y = 0\). To find the x-intercepts, set the quadratic equation to zero and solve for \(x\).
Using the quadratic formula: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \], we can solve our example:
For \( y = -x^2 - 4x - 3 \), where \( a = -1\), \( b = -4\), and \( c = -3 \):

• \(0 = -x^2 - 4x - 3\)
• \( x = \frac{4 \pm \sqrt{16 - 12}}{-2}\)
• \( x = \frac{4 \pm 2}{-2}\) \( x = \frac{6}{-2} = -3 \) and \( x = \frac{2}{-2} = -1 \)

Hence the x-intercepts are \((-3, 0)\) and \((-1, 0)\).

Domain and Range

Understanding the domain and range of a quadratic function is essential. The domain of a quadratic function is always all real numbers, as the x-values extend infinitely in both directions. Thus, for our example, the domain is \( (-\infty, \infty) \).

The range, however, is determined by the orientation and position of the parabola. Since the quadratic function \( y = -x^2 - 4x - 3 \) opens downwards (negative \(a\) value), and the vertex is at \( (2, -15) \), the vertex represents the highest point on the graph. Hence, the range includes all y-values less than or equal to -15.
This gives us the range: \( (-\infty, -15] \).