Since the square root of a negative number is not a real number, then a negative radical will result in no real solution.
Answer:
-111
Step-by-step explanation:
substitute 5 in place of each x
-4(5)^2 - 3(5) + 4
then evaluate or type it into your calculator
-100 - 15 + 4
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Look at the graph below carefully
Observe the results of shifting ={2}^{x}f(x)=2x
vertically:
The domain, (−∞,∞) remains unchanged.
When the function is shifted up 3 units to ={2}^{x}+3g(x)=2x +3:
The y-intercept shifts up 3 units to (0,4).
The asymptote shifts up 3 units to y=3y=3.
The range becomes (3,∞).
When the function is shifted down 3 units to ={2}^{x}-3h(x)=2 x −3:
The y-intercept shifts down 3 units to (0,−2).
The asymptote also shifts down 3 units to y=-3y=−3.
The range becomes (−3,∞).