Complete the square
factor out linear coefient
f(x)=7(x^2+6x)
take 1/2 of linear coefient and square it and add negative and positive to inside
6/2=3, 3^2=9
f(x)=7(x^2+6x+9-9)
complete squre
f(x)=7((x+3)^2-9)
expand
f(x)=7(x+3)^2-63 is veretx form
vertex is (-3,-63)
Pythagorean theorem. A^2+B^2=C^2.
So 12^2+16^2=c^2
144+256=c^2
400=c^2
square root of 400=c
So C=20
Hit the heart if it was helpful :D
16+3/7-(12+2/5)=16-12+3/7-2/5=4+(3*5-2*7)/35=4+1/35
Answer:
-3 1/3
Step-by-step explanation:
The quadratic
... y = ax² +bx +c
has its extreme value at
... x = -b/(2a)
Since a = 3 is positive, we know the parabola opens upward and the extreme value is a minimum. (We also know that from the problem statement asking us to find the minimum value.) The value of x at the minimum is -(-4)/(2·3) = 2/3.
To find the minimum value, we need to evaluate the function for x=2/3.
The most straightforward way to do this is to substitue 2/3 for x.
... y = 3(2/3)² -4(2/3) -2 = 3(4/9) -8/3 -2
... y = (4 -8 -6)/3 = -10/3
... y = -3 1/3
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<em>Confirmation</em>
You can also use a graphing calculator to show you the minimum.