1. `y=x^2+3x-4`, find Vertex of a function

Solution:
`y=x^2+3x-4`

1. Vertex :
`y=x^2+3x-4`

Method-1: Find vertex using polynomial form
Comparing the equation x^2+3x-4 with `ax^2+bx+c`, we get

`a=1,b=3,c=-4`

`h=(-b)/(2a)=(-3)/(2 * 1)=-3/2`

Now, substitute value of h in f(x), to find value of k
`k=f(h)=f(-3/2)=(-3/2)^2+3(-3/2)-4`

`:. k=1/2-9/2-4`

`:. k=-25/4`

Vertex `=(h,k)=(-3/2,-25/4)`

Method-2: Find vertex using vertex form `y=a(x-h)^2+k`

Completing the Square
`x^2+3x-4`

`=1 (x^2+3x-4)`

The coefficient of the x is `3`, so now we divide this by 2 : `(3 -: 2 = 3/2)`

and square it `(3/2)^2=9/4`. So we add and subtract `9/4`

`=1 (x^2+3x + 9/4 - 9/4 - 4)`

`=1 [(x^2+3x+9/4) -25/4]`

`=1[( x + 3/2 )^2 -25/4 ]`

`:. y=1(x-(-3/2))^2+(-25/4)`

Now compare with `y=a(x-h)^2+k`, we get

`a=1,h=-3/2,k=-25/4`

Vertex `=(h,k)=(-3/2,-25/4)`

If `a<0` then the vertex is a maximum value

If `a>0` then the vertex is a minimum value

Here `a=1>0`

So minimum Vertex = `(h,k)=(-3/2,-25/4)`

2. Graph :
some extra points to plot the graph
`y=f(x)=x^2+3x-4`

`f(-5)=(-5)^2+3(-5)-4=25-15-4=6`

`f(-4)=(-4)^2+3(-4)-4=16-12-4=0`

`f(-3)=(-3)^2+3(-3)-4=9-9-4=-4`

`f(-2)=(-2)^2+3(-2)-4=4-6-4=-6`

`f(-1)=(-1)^2+3(-1)-4=1-3-4=-6`

`f(0)=(0)^2+3(0)-4=0-4=-4`

`f(1)=(1)^2+3(1)-4=1+3-4=0`

`f(2)=(2)^2+3(2)-4=4+6-4=6`

`f(3)=(3)^2+3(3)-4=9+9-4=14`

graph