Question

g is a quadratic function with the vertex at ( - 1, 1) and g(0) = – 1. Find an algebraic equation for g(t).

Answer 1

`\begin{gathered} \text{ Substitute the vertex coordinate for h and k in the vertex form.} \\ \text{ Since vertex is }(-1,1),\text{ we substitute -1 for h and 1 for k} \\ y=a\mleft(x-h\mright)^2+k \\ y=a(x-(-1))^2+1 \\ y=a(x+1)^2+1 \\ \text{ Substitute }the\text{ point coordinates for x and y in the equation}.\text{ Given g(0)=-1, then} \\ \text{ we have the point (0,-1)},\text{ substituting x for 0 and y for -1},\text{ and solve for a} \\ y=a(x+1)^2+1 \\ -1=a(0+1)^2+1 \\ -1=a(1^2)+1 \\ -1=a+1 \\ -1-1=a \\ -2=a \\ a=-2 \\ \text{ Go back to the initial equation and substitute a} \\ y=-2(x+1)^2+1 \\ y=-2(x^2+2x+1)+1 \\ y=-2x^2-2x-2+1 \\ g(x)=-2x^2-2x-1 \\ \text{ Convert x to t} \\ g(t)=-2t^2-2t-1 \end{gathered}`