Respuesta :
Answer:
1) The zeros of the function is at t=-3,-1.
2) The vertex of the parabola is (-2,-1).
Step-by-step explanation:
Given : Function [tex]h(t)=t^2+4t+3[/tex]
To find :
1) What are the zeros of the function?
2) What is the vertex of the parabola?
Solution :
1) The zeros of the function,
To find the roots of the equation, replace h(t)=0 and solve.
[tex]t^2+4t+3=0[/tex]
Applying middle term split,
[tex]t^2+3t+t+3=0[/tex]
[tex]t(t+3)+1(t+3)=0[/tex]
[tex](t+3)(t+1)=0[/tex]
[tex]t=-3,-1[/tex]
The zeros of the function is at t=-3,-1.
2) The vertex of the parabola,
Comparing [tex]ax^2+bx+c[/tex] with [tex]t^2+4t+3[/tex]
Here, a=1, b=4, c=3
The x -coordinate of the vertex is given by [tex]-\frac{b}{2a}[/tex]
[tex]-\frac{b}{2a}=-\frac{4}{2(1)}=-2[/tex]
For y-coordinate put t=-2 in the function,
[tex]h(-2)=(-2)^2+4(-2)+3[/tex]
[tex]h(-2)=4-8+3[/tex]
[tex]h(-2)=-1[/tex]
The vertex of the parabola is (-2,-1).