Vertex of a function y=x^2+3x-4 Example-1

1. y=x2+3x-4 $y=x^2+3x-4$ , find Vertex of a function

Solution:

y=x2+3x-4 $y=x^2+3x-4$

1. Vertex :

y=x2+3x-4 $y=x^2+3x-4$

Method-1: Find vertex using polynomial form
Comparing the equation x^2+3x-4 with

ax2+bx+c $ax^2+bx+c$ , we get

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

h=-b2a=-32⋅1=-32 $h=(-b)/(2a)=(-3)/(2 * 1)=-3/2$

Now, substitute value of h in f(x), to find value of k

k=f(h)=f(-32)=(-32)2+3(-32)-4 $k=f(h)=f(-3/2)=(-3/2)^2+3(-3/2)-4$

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

$:. k=-25/4$

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

This material is intended as a summary. Use your textbook for detail explanation.
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