Question

find the coordinates for the vertex of the following quadratic function.
$f(x) = \frac{1}{2}(x + 1)(x + 5)$
show your work here

Explanation:

Step1: Find the x-intercepts

The quadratic function is given in factored form \( f(x)=\frac{1}{2}(x + 1)(x + 5) \). The x - intercepts occur when \( f(x)=0 \), so \( x+1 = 0\) or \(x + 5=0\). Solving these, we get \(x=-1\) and \(x=-5\).

Step2: Find the x - coordinate of the vertex

For a quadratic function in factored form \(y=a(x - r_1)(x - r_2)\), the x - coordinate of the vertex (the axis of symmetry) is the midpoint of the x - intercepts. The formula for the midpoint of two numbers \(x_1\) and \(x_2\) is \(x=\frac{x_1 + x_2}{2}\). Here, \(x_1=-1\) and \(x_2=-5\), so \(x=\frac{-1+(-5)}{2}=\frac{-6}{2}=-3\).

Step3: Find the y - coordinate of the vertex

Substitute \(x = - 3\) into the function \(f(x)=\frac{1}{2}(x + 1)(x + 5)\). First, calculate \((x + 1)\) and \((x + 5)\) when \(x=-3\): \(x + 1=-3 + 1=-2\) and \(x + 5=-3+5 = 2\). Then \(f(-3)=\frac{1}{2}\times(-2)\times(2)=\frac{1}{2}\times(-4)=-2\).

Answer:

The coordinates of the vertex are \((-3,-2)\)