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Need help in finding Vertex, simmetry and min/max
- Thread starter matteo70
- Start date
BigGlenntheHeavy
Senior Member
- Joined
- Mar 8, 2009
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\(\displaystyle Many \ ways \ to \ do \this, \ one \ way \ is \ expand, \ to \ wit:\)
\(\displaystyle f(x) \ = \ \frac{(x+3)^2}{4}+7 \ = \ \frac{x^2+6x+9}{4}+7 \ = \ \frac{x^2+6x+37}{4} \ = \ \frac{x^2}{4}+\frac{3x}{2}+\frac{37}{4}.\)
\(\displaystyle Now \ vertex \ = \ \frac{-b}{2a} \ = \ \frac{-3/2}{1/2} \ = \ -3\)
\(\displaystyle Hence \ x \ = \ -3 \ is \ your \ axis \ of \ symmetry \ and \ f(-3) \ = \ 7 \ is \ absolute \ min, \ (also \ vertex).\)
\(\displaystyle f(x) \ = \ \frac{(x+3)^2}{4}+7 \ = \ \frac{x^2+6x+9}{4}+7 \ = \ \frac{x^2+6x+37}{4} \ = \ \frac{x^2}{4}+\frac{3x}{2}+\frac{37}{4}.\)
\(\displaystyle Now \ vertex \ = \ \frac{-b}{2a} \ = \ \frac{-3/2}{1/2} \ = \ -3\)
\(\displaystyle Hence \ x \ = \ -3 \ is \ your \ axis \ of \ symmetry \ and \ f(-3) \ = \ 7 \ is \ absolute \ min, \ (also \ vertex).\)
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 \(\displaystyle a(x - h)^2 + k,\)
where \(\displaystyle (h, k)\) are the coordinates of the vertex.
Your equation can be rewritten as \(\displaystyle f(x) = \frac{1}{4}(x - (-3))^2 + 7\)
So, the coordinates of the vertex are \(\displaystyle (-3, 7).\)
Because the coefficient on the \(\displaystyle x^2 \ term\) 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 \(\displaystyle x = -3,\) which is a direct connection to the vertex \(\displaystyle (-3, 7).\)