I am stuck on finding the above mentioned for the f(x)=1/4(x+3)^2+7 Can anyone help me please?
M matteo70 New member Joined Oct 22, 2010 Messages 2 Oct 22, 2010 #1 I am stuck on finding the above mentioned for the f(x)=1/4(x+3)^2+7 Can anyone help me please?
B BigGlenntheHeavy Senior Member Joined Mar 8, 2009 Messages 1,577 Oct 22, 2010 #2 \(\displaystyle Many \ ways \ to \ do \this, \ one \ way \ is \ expand, \ to \ wit:\) 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}.f(x) = 4(x+3)2+7 = 4x2+6x+9+7 = 4x2+6x+37 = 4x2+23x+437. Now vertex = −b2a = −3/21/2 = −3\displaystyle Now \ vertex \ = \ \frac{-b}{2a} \ = \ \frac{-3/2}{1/2} \ = \ -3Now vertex = 2a−b = 1/2−3/2 = −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).Hence x = −3 is your axis of symmetry and f(−3) = 7 is absolute min, (also vertex).
\(\displaystyle Many \ ways \ to \ do \this, \ one \ way \ is \ expand, \ to \ wit:\) 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}.f(x) = 4(x+3)2+7 = 4x2+6x+9+7 = 4x2+6x+37 = 4x2+23x+437. Now vertex = −b2a = −3/21/2 = −3\displaystyle Now \ vertex \ = \ \frac{-b}{2a} \ = \ \frac{-3/2}{1/2} \ = \ -3Now vertex = 2a−b = 1/2−3/2 = −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).Hence x = −3 is your axis of symmetry and f(−3) = 7 is absolute min, (also vertex).
L lookagain Elite Member Joined Aug 22, 2010 Messages 3,262 Oct 22, 2010 #4 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? Click to expand... matteo70, this form lends itself to the following, because it is of the form a(x−h)2+k,\displaystyle a(x - h)^2 + k,a(x−h)2+k, where (h,k)\displaystyle (h, k)(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 + 7f(x)=41(x−(−3))2+7 So, the coordinates of the vertex are (−3,7).\displaystyle (-3, 7).(−3,7). Because the coefficient on the x2 term\displaystyle x^2 \ termx2 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,x=−3, which is a direct connection to the vertex (−3,7).\displaystyle (-3, 7).(−3,7).
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? Click to expand... matteo70, this form lends itself to the following, because it is of the form a(x−h)2+k,\displaystyle a(x - h)^2 + k,a(x−h)2+k, where (h,k)\displaystyle (h, k)(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 + 7f(x)=41(x−(−3))2+7 So, the coordinates of the vertex are (−3,7).\displaystyle (-3, 7).(−3,7). Because the coefficient on the x2 term\displaystyle x^2 \ termx2 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,x=−3, which is a direct connection to the vertex (−3,7).\displaystyle (-3, 7).(−3,7).