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:

$\displaystyle t=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

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:

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

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:

$y=t^2+19t+(\frac{19}{2})^2-(\frac{19}{2})^2+60$

Now we apply the identity:

$y=(t-\frac{19}{2})^2-\frac{361}{4}+60=(t-\frac{19}{2})^2-\frac{121}{4}$

Finally, we add 121/4:

$y+\frac{121}{4}=(t-\frac{19}{2})^2$

Comparing with the vertex form of the quadratic function:

a=1

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