To do this problem, you need to use a process called completing the square. Let me explain:
To complete the square on the function f(x) = x² + 8x +13, first group the first two terms in ( ) and leave some space at the end as follows:
f(x) = (x² + 8x ) + 13 Now our next step is to fill in the space and adjust our expression on the right hand side of the function. To do this, we take half of the middle number 8 and then square it: so 4² = 16 and we fill in our space inside the ( ) with this value 16;
f(x) = (x² + 8x + 16) + 13 now what we have done is to increase the overall value of our expression on the right by 16, but we want the overall value to remain the same. To fix this we simply need to subtract 16 at the end like this: f(x) = (x² + 8x + 16) + 13 -16 we can simplify and get the following.
f(x) = (x² + 8x + 16) - 3 At this point we're almost done.. All we need to do now is to rewrite the what is in the parentheses in a slightly different form. Here is what it will look like: f(x) = (x + 4)² - 3 notice all I did was take the sum of the square root of x² and the square root of 16 originally in the ( ) to get then new expression inside the ( ) and then square that ( )²
Now this is a nice form to have because you can get the vertex straight from this form.. IN FACT this is called vertex form or (h,k) form for short. In general the form is f(x) = a(x - h)² + k don't worry about the 'a' for now.. you might see that in our case it is just 1 and will not effect our equation. You only have to consider this if the original leading coefficient of the quadratic is not 1 to begin with...
So you can see that our vertex is (-4,-3)
Hope this is helpful, but if you have questions let me know.
Answer:
y = (5/4)2^x
Step-by-step explanation:
The function value increases by a factor of 40/10 = 4 when x increases by 2. The function can be written as ...
y = (reference value)·(growth factor)^((x -reference)/(change in x for growth factor))
y = 10·4^((x-3)/2) . . . . . . using point (3, 10) as a reference
This can be simplified to ...
y = 10·2^(x -3) = 10/8·2^x
y = (5/4)2^x
If the slope of AB = CD and BC = AD it's a parallelogram:
Slope of AB = 6+1 / -9+5 = -7/4
CD = -2-5 / 3+1 = -74
These are equal.
BC = 5-6 / -1 +9 = -1/8
AD = -2 +1 / 3+5 = -1/8
These are also equal so it is a parallelogram.
Now to find if the diagonals are perpendicular find the slope of the perpensicular points:
AC = 5 +1 / -1 +5 = 6/4 = 3/2
BD = 6+2 / -9 -3 = 8/-12 = -2/3
Because BD is the reciprocal of AC, this means they are perpendicular.
And because AB is not perpendicular to AD ( AB and AD are not reciprocals) it is a rhombus.
The food service gave out 80 samples, which is 4 fewer than 70% of the total number of samples they prepared. Lets make x the total number of samples they prepared.
So lets write it as an equation
80 = (70% of x) -4
70% converts to a decimal by moving the decimal point 2 spaces up, so it is 0.7
80 = 0.7x-4
Add 4 to both sides of the equation
84 = 0.7x
Divide both sides by 0.7 to isolate the x
x = 120
The food service prepared 120 samples.
