Refresher..
Positive graph.. Vertex- (-1,0)
f(x)=(x-h)^2 +k f(x)= a(x-(-1))^2 +0 = a(x+1)^2 +0
Not sure if I did that right so far..
yes. and to find a, use a point on the curve (but not the vertex) for example (0,1)
1= a(0+1)^2 1 = a so a is 1 and the equation is y= (x+1)^2 or multiplied out y= x^2 +2x+1
or you could test the point (-1,0) in each of the choices that works for choices B and C then use (0,1) to see C is the only equation that works
True.
another way: use -b/(2a) to find the x value of the vertex you want that to be -1 and only choice C works
Hmm never thought of that
the first two choices have b=0 so you get x=0. those are wrong. choice C gives -2/2= -1 (that works) choice D give s -(-2)/2= +1 (nope)
Okay.
Next problem..
Find the range of the quadratic function. f(x)= x^2 -8x-5
parabola with \( \cup\) shape so from the y value of the vertex to + infinity
Um do you mean -5 to infinity
Sorry I dont really remember how to solve this one.
range are the y values it is U shaped, so you want the smallest y value (which happens at the vertex)
you find the x value of the vertex using -b/(2a) and use that x value to find the corresponding y value. that will be the smallest y value on the curve. and of course y will go to + infinity as you put in bigger and bigger x values
Okay thats what I thought
but the smallest y is not -5
Yeah sorry I was confused at first
Vertex- (4,-12) Range- [-21, infinity)
yes, looks good
Okay
A rain gutter is made from sheets of aluminum that are 18 inches wide by turning up the edges to form right angles. Determine the depth of the gutter that will maximize its cross-sectional area and allow the greatest amount of water to flow.
A= 18x -2x^2
x= -(-2) over 2(18) .... I think I need to change it to -2x^2 +18
Answer -> 4.5
A= 18x -2x^2 yes, but generally we write it in standard form A= -2x^2 + 18x this is a \(\cap\) shape, and its vertex has the biggest y value (in this case largest area)
x of vertex is -b/(2a)= -18/(2*-2) = 18/4 = 9/2 = 4.5 yes you found the correct answer: d= 4.5 inches to maximizes the area of the cross-section
Okay thank you.
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