Bytelearn - cat image with glassesAI tutor

Welcome to Bytelearn!

Let’s check out your problem:

Paul opened a bakery. The net value of the bakery (in thousands of dollars) tt months after its creation is modeled by Paul wants to know what his bakery's lowest net value will be v(t)=2t212t141)v(t)=2t^2-12t-141). Rewrite the function in a different form (factored or vertex) where the answer appears as a number in the equation. v(t)=v(t)=.What is the bakeries lowest net value?

Full solution

Q. Paul opened a bakery. The net value of the bakery (in thousands of dollars) tt months after its creation is modeled by Paul wants to know what his bakery's lowest net value will be v(t)=2t212t141)v(t)=2t^2-12t-141). Rewrite the function in a different form (factored or vertex) where the answer appears as a number in the equation. v(t)=v(t)=.What is the bakeries lowest net value?
  1. Rewrite Quadratic Function: First, let's rewrite the quadratic function v(t)=2t212t141v(t) = 2t^2 - 12t - 141 in vertex form to find the minimum value.
  2. Vertex Form Definition: The vertex form of a quadratic function is v(t)=a(th)2+kv(t) = a(t-h)^2 + k, where (h,k)(h, k) is the vertex of the parabola.
  3. Calculate Vertex Coordinates: To find hh, use the formula h=b2ah = -\frac{b}{2a}. Here, a=2a = 2 and b=12b = -12. So, h=122×2=3h = -\frac{-12}{2 \times 2} = 3.
  4. Substitute to Find Minimum Value: Substitute t=3t = 3 back into the original equation to find kk. v(3)=2(3)212(3)141=1836141=159v(3) = 2(3)^2 - 12(3) - 141 = 18 - 36 - 141 = -159.

More problems from Interpreting Linear Expressions