Question

H(t)=t² + 4t + 3

Answer 1

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 $h(t)=t^2+4t+3$

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.

$t^2+4t+3=0$

Applying middle term split,

$t^2+3t+t+3=0$

$t(t+3)+1(t+3)=0$

$(t+3)(t+1)=0$

$t=-3,-1$

The zeros of the function is at t=-3,-1.

2) The vertex of the parabola,

Comparing $ax^2+bx+c$ with $t^2+4t+3$

Here, a=1, b=4, c=3

The x -coordinate of the vertex is given by $-\frac{b}{2a}$

$-\frac{b}{2a}=-\frac{4}{2(1)}=-2$

For y-coordinate put t=-2 in the function,

$h(-2)=(-2)^2+4(-2)+3$

$h(-2)=4-8+3$

$h(-2)=-1$

The vertex of the parabola is (-2,-1).