Need help in finding Vertex, simmetry and min/max

matteo70

New member
Joined
Oct 22, 2010
Messages
2
I am stuck on finding the above mentioned for the f(x)=1/4(x+3)^2+7
Can anyone help me please?
 
\(\displaystyle Many \ ways \ to \ do \this, \ one \ way \ is \ expand, \ to \ wit:\)

f(x) = (x+3)24+7 = x2+6x+94+7 = x2+6x+374 = x24+3x2+374.\displaystyle f(x) \ = \ \frac{(x+3)^2}{4}+7 \ = \ \frac{x^2+6x+9}{4}+7 \ = \ \frac{x^2+6x+37}{4} \ = \ \frac{x^2}{4}+\frac{3x}{2}+\frac{37}{4}.

Now vertex = b2a = 3/21/2 = 3\displaystyle Now \ vertex \ = \ \frac{-b}{2a} \ = \ \frac{-3/2}{1/2} \ = \ -3

Hence x = 3 is your axis of symmetry and f(3) = 7 is absolute min, (also vertex).\displaystyle Hence \ x \ = \ -3 \ is \ your \ axis \ of \ symmetry \ and \ f(-3) \ = \ 7 \ is \ absolute \ min, \ (also \ vertex).
 
matteo70 said:
I am stuck on finding the above mentioned for the f(x)=1/4(x+3)^2+7
Can anyone help me please?

matteo70,

this form lends itself to the following, because it is of the form a(xh)2+k,\displaystyle a(x - h)^2 + k,
where (h,k)\displaystyle (h, k) are the coordinates of the vertex.

Your equation can be rewritten as f(x)=14(x(3))2+7\displaystyle f(x) = \frac{1}{4}(x - (-3))^2 + 7

So, the coordinates of the vertex are (3,7).\displaystyle (-3, 7).

Because the coefficient on the x2 term\displaystyle x^2 \ term is positive, the parabola opens up and has a minimum point.

Then the vertex is the minimum point.

The parabola is symmetrical about the line x=3,\displaystyle x = -3, which is a direct connection to the vertex (3,7).\displaystyle (-3, 7).
 
Top