The distance would be 9.
Hope I could help.
1)
i) Reflection about y axis i.e. x=0
ii) Horizontal shrink by scale factor 4
iii) vertical shift by 4 units up.
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2) f(x) passes through (1,-1) and (0,1)
So slope = (1+1)/(0-1) = -2
Using point slope form equation of f(x) is
y-1 = -2 (x-0) or y = -2x+1
g(x) passes through (0,-1) and (1,1)
Slope = (2/1) = 2
So using point slope form g(x) is
y-1=2x
Hence we find that y-1 = -2x is transformed into y-1 = 2x
i.e. there is a reflection of f(x) on the line y =1
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A quadratic has the form ax^2+bx+c
To factor a quadratic you want to find two values, j and k, which satisfy two conditions...
j*k=a*c and j+k=b
In this case ac=-140 and b=4.
The only values possible are j=14 and k=-10
so jk=-140 and j+k=4 which are the conditions we needed to satisfy...
Now you remove the linear, or "b term", from the original equation and replace it with jx and kx... and in this case you will have:
-2h^2-14h+10h+70 now you can factor out the greatest common factor from the first pair of terms and the second pair of terms...
-2h(h+7)+10(h+7) which is equal to
(-2h+10)(h+7) and you can factor the first term again if you wish...
-2(h-5)(h+7)
<h3>
Answer: y = x+1</h3>
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Explanation:
f(x) = x^3 - 2x + 3
f ' (x) = 3x^2 - 2 ..... apply the power rule
f ' (1) = 3(1)^2 - 2 ... plug in x coordinate of given point
f ' (1) = 1
If x = 1 is plugged into the derivative function, then we get the output 1. This means the slope of the tangent line at (1,2) is m = 1. It's just a coincidence that the x input value is the same as the slope m value.
Now apply point slope form to find the equation of the tangent line
y - y1 = m(x - x1)
y - 2 = 1(x - 1)
y - 2 = x - 1
y = x - 1 + 2
y = x + 1 is the equation of the tangent line.
The graph is shown below. I used GeoGebra to make the graph.
