Paul opened a bakery. The net value of the bakery (in thousands of dollars)

Question

\[t\] months after its creation is modeled by
\[v(t)=2t^2-12t-14\]
Paul wants to know what his bakery's lowest net value will be.
1) Rewrite the function in a different form (factored or vertex) where the answer appears as a number in the equation.
\[v(t)=\]
2) What is the bakery's lowest net value?
thousand dollars

Answer 1

To find the lowest net value of Paul's bakery, we need to work with the quadratic function given by:

\[ v(t) = 2t^2 - 12t - 14 \]

1. First, we can rewrite this function in vertex form. The vertex form of a quadratic function can be represented as:

\[ v(t) = a(t - h)^2 + k \]

where \((h, k)\) is the vertex of the parabola. To do this, we can complete the square.

Starting with the equation:

\[ v(t) = 2t^2 - 12t - 14 \]

First, factor out the coefficient of \(t^2\) from the first two terms:

\[ v(t) = 2(t^2 - 6t) - 14 \]

Next, to complete the square inside the parentheses, take half of the coefficient of \(t\) (which is -6), square it, and add and subtract that inside the parentheses:

Half of -6 is -3, and \((-3)^2 = 9\). So we can rewrite:

\[ v(t) = 2(t^2 - 6t + 9 - 9) - 14 \]

This simplifies to:

\[ v(t) = 2((t - 3)^2 - 9) - 14 \] \[ = 2(t - 3)^2 - 18 - 14 \] \[ = 2(t - 3)^2 - 32 \]

So the function in vertex form is:

\[ v(t) = 2(t - 3)^2 - 32 \]

2. The vertex of the parabola represented by this equation occurs at \(t = 3\), and the value of \(v(t)\) at this vertex is the minimum value of the function:

The lowest net value of the bakery is:

\[ -32 \text{ (in thousands of dollars)} \]

Therefore, the bakery's lowest net value is:

\(-32\) thousand dollars.