Respuesta :
Quadratic Function
We are given the function
[tex]f\mleft(t\mright)=t^2+19t+60[/tex]1) Find the zeros of the function
The zeros or roots of f, are found by equating it to 0:
[tex]t^2+19t+60=0[/tex]The standard representation of a quadratic function is:
[tex]f\mleft(t\mright)=at^2+bt+c[/tex]where a,b, and c are constants.
Solving with the quadratic formula:
[tex]\displaystyle t=\frac{-b\pm\sqrt{b^2-4ac}}{2a}[/tex]The coefficients can be established by comparing the generic equation with the given equation, thus: a=1, b=19, c=60. Substituting:
[tex]\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}[/tex]Operating:
[tex]\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}[/tex]The smaller root is -15, the larger root is -4. Answer:
-15
-4
2) The quadratic equation can be also written in vertex form:
[tex]y-k=a\mleft(x-h\mright)^2[/tex]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:
[tex]\mleft(a+b\mright)^2=a^2+2ab+b^2[/tex]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:
[tex]y=t^2+19t+(\frac{19}{2})^2-(\frac{19}{2})^2+60[/tex]Now we apply the identity:
[tex]y=(t-\frac{19}{2})^2-\frac{361}{4}+60=(t-\frac{19}{2})^2-\frac{121}{4}[/tex]Finally, we add 121/4:
[tex]y+\frac{121}{4}=(t-\frac{19}{2})^2[/tex]Comparing with the vertex form of the quadratic function:
a=1
Vertex: (19/2,-121/4)
