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I think this is a simple question but i do not know how...hopefully you guys here could clear my doubt..

The question is:

>>> Given that the equation of a curve is y = x^2 + 3x - 4, tell me how to find the value of the minimum point of the curve?

Please show me the steps..I really want to understand..thanks.

2007-10-07 19:33:23 · 6 answers · asked by Anonymous in Science & Mathematics Mathematics

6 answers

Are you in calculus or not? If you are, differentiate and find when it's 0.

dy/dx = 2x +3 = 0.
x = -3/2


If you're in pre-cal or some other course, use the non-descriminate part of the quadratic equation.
x = -b/2a.
b = 3
a = 1.
Bam.

Isn't calculus easier?

2007-10-07 19:37:53 · answer #1 · answered by Chase 3 · 1 0

this method is apply only for quadratic form


let y= ax^2 + bx + c

if a>0 then this graph has minimum point

where x=-b/2a, substitute x into the equation to get y value



if a <0, this graph has a maximum point

where x=-b/2a, y from the equation


for thius question

a=1 b=3 c=-4

a>0, has minimum value where x = - 3/2(1)
for y, substitute x into the equation by your self.
you hav to practice more

2007-10-08 03:11:27 · answer #2 · answered by pierre m 2 · 0 0

Using algebra,
y = x^2 + 3x - 4
y = x^2 + 3x + (3/2)^2 - (3/2)^2 - 4
y = (x + 3/2)^2 - 9/4 - 4
y = (x + 3/2)^2 - 9/4 - 16/4
y = (x + 3/2)^2 - 25/4
The minimum point is at (- 3/2, -25/4)

Using calculus,
y = x^2 + 3x - 4
y' = 2x + 3 = 0
x = - 3/2
y = (- 3/2)^2 + 3(- 3/2) - 4
y = 9/4 - 9/2 - 4
y = (9 - 18 - 16)/4
y = - 25/4
Minimum point:
(- 3/2, - 25/4)

2007-10-08 02:50:37 · answer #3 · answered by Helmut 7 · 0 0

y = x^2 + 3x - 4

Finding max or min? Do derivative

y' = 2x +3

set y' = 0

0 = 2x +3

x= -3/2 This is the only value and the minimum.

2007-10-08 02:39:06 · answer #4 · answered by Romuald 2 · 0 0

minimze y
y'=2x+3
y'=0
2x+3=0
x=-3/2
y=(-3/2)^2+3(-3/2)-4=-6.25

therefore at P(-3/2,-6.25) y is minimum

to prove that y is minimum at that point
let us solve for the second derivative of y
for y to be minimum y'' must be positive
y"=2

Also using analytic geometry it could be seen
that the coefficient of x^2 is greater
than zero therefore the parabola opens upward

2007-10-08 02:42:55 · answer #5 · answered by ptolemy862000 4 · 0 0

y=x^2+3x-4
y=(x+4)(x-1)
when y=0 x=-4 andx=1
when x is half way between -4 and 1 y will be at a minimum
substitute this number into the equation
y=(-1.5)^2+3(-1.5)-4
y=2.25-4.5-4
y=-6.25 and x=-1.5

2007-10-08 02:54:47 · answer #6 · answered by john 4 · 0 0

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