Question

Suppose that \(f(x)=x^{2}+2 x-8\) (a) What is the vertex of \(f ?\) (b) What are the \(x\) -intercepts of the graph of \(f ?\) (c) Solve \(f(x)=-8\) for \(x\). What points are on the graph of \(f ?\) (d) Use the information obtained in parts (a)-(c) to \(\operatorname{graph} f(x)=x^{2}+2 x-8\)

Short answer

Vertex: (-1, -9). X-intercepts: 2 and -4. Points for \(f(x) = -8\): (0, -8) and (-2, -8).

Step-by-step solution

Find the vertex of the function

The vertex of a quadratic function in the form of \[ f(x) = ax^2 + bx + c \]can be found using the formula for the x-coordinate of the vertex: \[ x = -\frac{b}{2a} \]For the given function, \(f(x) = x^2 + 2x - 8\), where \(a = 1\) and \(b = 2\). Substituting these values, we get the x-coordinate as: \[ x = -\frac{2}{2(1)} = -1 \]To find the y-coordinate, substitute \(x = -1\) back into the function: \[ f(-1) = (-1)^2 + 2(-1) - 8 = 1 - 2 - 8 = -9 \]So, the vertex is \((-1, -9)\).

Find the x-intercepts

The x-intercepts occur where the function \(f(x)\) is equal to 0. Solve the equation \[ x^2 + 2x - 8 = 0 \]Use the quadratic formula, \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]Substitute \(a = 1\), \(b = 2\), and \(c = -8\), then calculate: \[ x = \frac{-2 \pm \sqrt{2^2 - 4(1)(-8)}}{2(1)} = \frac{-2 \pm \sqrt{4 + 32}}{2} = \frac{-2 \pm 6}{2} \]This gives us two solutions: \[ x = \frac{-2 + 6}{2} = 2 \]\[ x = \frac{-2 - 6}{2} = -4 \]So, the x-intercepts are \(2\) and \(-4\).

Solve \(f(x) = -8\)

To solve for \(f(x) = -8\), set the function equal to \(-8\): \[ x^2 + 2x - 8 = -8 \]Adding 8 to both sides gives: \[ x^2 + 2x = 0 \]Factor out the common term: \[ x(x + 2) = 0 \]Set each factor to zero: \[ x = 0 \]\[ x + 2 = 0 \Rightarrow x = -2 \]Thus, two points on the graph are \((0, -8)\) and \((-2, -8)\).

Graph the function

Using the vertex \((-1, -9)\), x-intercepts \(2\) and \(-4\), and points from \(f(x) = -8\) at \((0, -8)\) and \((-2, -8)\), plot these points on a coordinate plane. Sketch a parabola opening upwards as it is a quadratic function with a positive leading coefficient \(a=1\).

Key concepts

vertex of a parabola

Understanding the vertex of a parabola is essential for solving quadratic functions. The vertex represents the peak or the lowest point on the graph, depending on the direction the parabola opens. For the quadratic function in the form \( f(x) = ax^2 + bx + c \), the x-coordinate of the vertex is determined using the formula: \[ x = -\frac{b}{2a} \]
Once you find the x-coordinate, you can find the y-coordinate by substituting the x value back into the original function. This gives you the vertex as a coordinate pair \((x, y)\). For example, with the function \( f(x) = x^2 + 2x - 8 \), we find the vertex as follows:

• Calculate x: \( x = -\frac{2}{2(1)} = -1 \)
• Substitute \( x = -1 \) into \( f(x) \) to find y: \( f(-1) = (-1)^2 + 2(-1) - 8 = -9 \)

Thus, the vertex is \((-1, -9)\). Knowing the vertex gives us a better understanding of the graph's shape and position.

x-intercepts

The x-intercepts are the points where the graph of a quadratic function crosses the x-axis. This occurs when \( f(x) = 0 \). Finding the x-intercepts involves solving the equation \( ax^2 + bx + c = 0 \) using the quadratic formula: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Substituting the coefficients from \( f(x) = x^2 + 2x - 8 \) into the quadratic formula, we get:

• \( a = 1, b = 2, c = -8 \)
• Plug values into formula: \( x = \frac{-2 \pm \sqrt{4 + 32}}{2} = \frac{-2 \pm 6}{2} \)
• Solve for x: \( x = 2 \) and \( x = -4 \)

So, the x-intercepts for the function are at \( x = 2 \) and \( x = -4 \). These points tell us where the graph touches or crosses the x-axis.

solving quadratic equations

Solving quadratic equations involves finding the values of x that satisfy the equation \( ax^2 + bx + c = 0 \). One way to do this is by using factoring, completing the square, or the quadratic formula. For the function \( f(x) = x^2 + 2x - 8 \), we can solve \( f(x) = -8 \) as follows:

• Set \( f(x) = -8 \): \( x^2 + 2x - 8 = -8 \)
• Simplify: \( x^2 + 2x = 0 \)
• Factor: \( x(x + 2) = 0 \)
• Set each factor to zero: \( x = 0 \) and \( x + 2 = 0 \rightarrow x = -2 \)

Thus, the solutions are \( x = 0 \) and \( x = -2 \). These x values correspond to points on the graph where the function equals -8, specifically at \( (0, -8) \) and \( (-2, -8) \). This process highlights how to find specific points on the graph that meet certain conditions.

factoring polynomials

Factoring polynomials is a useful technique for solving quadratic equations, simplifying expressions, and understanding the structure of polynomial functions. A polynomial like \( x^2 + 2x - 8 \) can often be factored into the product of two binomials. To factor the given quadratic function, follow these steps:

• Look for two numbers that multiply to \( -8 \) (the constant term, \( c \)) and add to \( 2 \) (the linear coefficient, \( b \))
• The numbers \( 4 \) and \( -2 \) fit this criterion since \( 4 \times (-2) = -8 \) and \( 4 + (-2) = 2 \)
• Write the function as: \( x^2 + 2x - 8 = (x + 4)(x - 2) \)

Factoring simplifies solving the equation \( x^2 + 2x - 8 = 0 \), as we can now solve for x in \( (x + 4)(x - 2) = 0 \). Setting each factor to zero gives \( x = -4 \) and \( x = 2 \). Understanding how to factor quadratics is key to mastering polynomial equations.