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Andre45 [30]
Answer:
the one real zero is in the interval (-1, 0)
Step-by-step explanation:
Descartes' rule of signs tells you there are 0 or 2 positive real zeros. Changing the signs of the odd-degree terms and applying that rule again tells you there is one negative real zero. At the same time, those coefficients (-3, -5, -5, +7) have a negative sum, so you know ...
f(-1) = -6
f(0) = +7
so there is a zero in the interval (-1, 0).
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You can try a few values between x=0 and x=10 to see what the function does in that part of the graph. You find ...
f(1) = 10
f(2) = 21
f(3) = 58
So, it is safe to conclude that there are no real zeros for x > 0.
The only real zero of f(x) is in the interval (-1, 0).
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I like to use a graphing calculator for problems like this.
Answer: y = 2/3x + 3
Step-by-step explanation:
We will write the equation in slope intercept form (y=mx+b).
In the formula y=mx + b, m is the slope and b is the y intercept. We need the slope and y intercept numbers to write the equation. Since we know the slope we will use the given point to find the y intercept by plotting in the x and y coordinates into the formula.
1 = 2/3(-3) + b Solve for b
1= -6/3 + b
1 = -2 + b
+2 +2
b = 3
The equation will be y = 2/3x + 3