Consider the function 
First, factor it:
The x-intercepts are at points 
1. From the attached graph you can see that
- function is positive for
- function is negative for
2. Since
the function is neither even nor odd.
3. The domain is 
the range is 
Slope intercept form looks like:
y = mx + b
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Choose the one that follows the formula
B) y=-2x+15 is your answer.
-2 = m
15 = b
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hope this helps
Answer:
Roots are not real
Step-by-step explanation:
To prove : The roots of x^2 +(1-k)x+k-3=0x
2
+(1−k)x+k−3=0 are real for all real values of k ?
Solution :
The roots are real when discriminant is greater than equal to zero.
i.e. b^2-4ac\geq 0b
2
−4ac≥0
The quadratic equation x^2 +(1-k)x+k-3=0x
2
+(1−k)x+k−3=0
Here, a=1, b=1-k and c=k-3
Substitute the values,
We find the discriminant,
D=(1-k)^2-4(1)(k-3)D=(1−k)
2
−4(1)(k−3)
D=1+k^2-2k-4k+12D=1+k
2
−2k−4k+12
D=k^2-6k+13D=k
2
−6k+13
D=(k-(3+2i))(k+(3+2i))D=(k−(3+2i))(k+(3+2i))
For roots to be real, D ≥ 0
But the roots are imaginary therefore the roots of the given equation are not real for any value of k.
Answer:
Step-by-step explanation:
We are asked to find the tangent line approximation for 
near 
.
We will use linear approximation formula for a tangent line 
of a function 
at 
to solve our given problem.
Let us find value of function at as:
Now, we will find derivative of given function as:
Let us find derivative at
Upon substituting our given values in linear approximation formula, we will get:
Therefore, our required tangent line for approximation would be .
