Question

f(t) = t^2 + 19t + 601) What are the zeros of the function?Write the smaller t first, and the larger t second.smaller tlarger t2) What is the verte of the parabola?

Answer 1

Quadratic Function

We are given the function

`f\mleft(t\mright)=t^2+19t+60`

1) Find the zeros of the function

The zeros or roots of f, are found by equating it to 0:

`t^2+19t+60=0`

The standard representation of a quadratic function is:

`f\mleft(t\mright)=at^2+bt+c`

where a,b, and c are constants.

Solving with the quadratic formula:

`\displaystyle t=(-b\pm√(b^2-4ac))/(2a)`

The coefficients can be established by comparing the generic equation with the given equation, thus: a=1, b=19, c=60. Substituting:

`\displaystyle t=\frac{-19\pm\sqrt[]{19^2-4(1)(60)}}{2(1)}=\frac{-19\pm\sqrt[]{361-240}}{2}=\frac{-19\pm\sqrt[]{121}}{2}`

Operating:

`\begin{gathered} t=(-19\pm11)/(2) \\ \text{There are two solutions:} \\ t=(-19+11)/(2)=-(8)/(2)=-4 \\ t=(-19-11)/(2)=-(30)/(2)=-15 \end{gathered}`

The smaller root is -15, the larger root is -4. Answer:

-15

-4

2) The quadratic equation can be also written in vertex form:

`y-k=a\mleft(x-h\mright)^2`

Where a is the leading coefficient and (h,k) is the vertex.

To find this form, we need to complete squares as follows.

We need to recall this identity:

`\mleft(a+b\mright)^2=a^2+2ab+b^2`

Comparing the given function, we can see the first term is t=a, the second term should be 2ab=19t, thus b=19/2

To complete the square, we need to add and subtract b^2 as follows:

`y=t^2+19t+((19)/(2))^2-((19)/(2))^2+60`

Now we apply the identity:

`y=(t-(19)/(2))^2-(361)/(4)+60=(t-(19)/(2))^2-(121)/(4)`

Finally, we add 121/4:

`y+(121)/(4)=(t-(19)/(2))^2`

Comparing with the vertex form of the quadratic function:

a=1

Vertex: (19/2,-121/4)