The product of two positive integers plus their sum is 95. The integers are relatively prime, and each is less than 20. What is
1 answer:
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1. f(x)=x²+10x+16
Use the formula to find the vertex = (-b/2a, f(-b/2a)) , here in the above equation a=1(As, a>0 the parabola is open upward), b=10. by putting the values.
-b/2a = -10/2(1) = -5
f(-b/2a)= f(-5)= (-5)²+10(-5)+16= -9
So, Vertex = (-5, -9)
Now, find y- intercept put x=0 in the above equation. f(0)= 0+0+16, we get point (0,16).
Now find x-intercept put y=0 in the above equation. 0= x²+10x+16
x²+10x+16=0 ⇒x²+8x+2x+16=0 ⇒x(x+8)+2(x+8)=0 ⇒(x+8)(x+2)=0 ⇒x=-8 , x=-2
From vertex, y-intercept and x-intercept you can easily plot the graph of given parabolic equation. The graph is attached below.
2. f(x)=−(x−3)(x+1)
By multiplying the factors, the general form is f(x)= -x²+2x+3.
Use the formula to find the vertex = (-b/2a, f(-b/2a)) , here in the above equation a=-1(As, a<0 the parabola is open downward), b=2. by putting the values.
-b/2a = -2/2(-1) = 1
f(-b/2a)= f(1)=-(1)²+2(1)+3= 4
So, Vertex = (1, 4)
Now, find y- intercept put x=0 in the above equation. f(0)= 0+0+3, we get point (0, 3).
Now find x-intercept put y=0 in the above equation. 0= -x²+2x+3.
-x²+2x+3=0 the factor form is already given in the question so, ⇒-(x-3)(x+1)=0 ⇒x=3 , x=-1
From vertex, y-intercept and x-intercept you can easily plot the graph of given parabolic equation. The graph is attached below.
3. f(x)= −x²+4
Use the formula to find the vertex = (-b/2a, f(-b/2a)) , here in the above equation a=-1(As, a<0 the parabola is open downward), b=0. by putting the values.
-b/2a = -0/2(-1) = 0
f(-b/2a)= f(0)= −(0)²+4 =4
So, Vertex = (0, 4)
Now, find y- intercept put x=0 in the above equation. f(0)= −(0)²+4, we get point (0, 4).
Now find x-intercept put y=0 in the above equation. 0= −x²+4
−x²+4=0 ⇒-(x²-4)=0 ⇒ -(x-2)(x+2)=0 ⇒x=2 , x=-2
From vertex, y-intercept and x-intercept you can easily plot the graph of given parabolic equation. The graph is attached below.
4. f(x)=2x²+16x+30
Use the formula to find the vertex = (-b/2a, f(-b/2a)) , here in the above equation a=2(As, a>0 the parabola is open upward), b=16. by putting the values.
-b/2a = -16/2(2) = -4
f(-b/2a)= f(-4)= 2(-4)²+16(-4)+30 = -2
So, Vertex = (-4, -2)
Now, find y- intercept put x=0 in the above equation. f(0)= 0+0+30, we get point (0, 30).
Now find x-intercept put y=0 in the above equation. 0=2x²+16x+30
2x²+16x+30=0 ⇒2(x²+8x+15)=0 ⇒x²+8x+15=0 ⇒x²+5x+3x+15=0 ⇒x(x+5)+3(x+5)=0 ⇒(x+5)(x+3)=0 ⇒x=-5 , x= -3
From vertex, y-intercept and x-intercept you can easily plot the graph of given parabolic equation. The graph is attached below.
5. y=(x+2)²+4
The general form of parabola is y=a(x-h)²+k , where vertex = (h,k)
if a>0 parabola is opened upward.
if a<0 parabola is opened downward.
Compare the given equation with general form of parabola.
-h=2 ⇒h=-2
k=4
so, vertex= (-2, 4)
As, a=1 which is greater than 0 so parabola is opened upward and the graph has minimum.
The graph is attached below.
Answer:
It is continuous since
Step-by-step explanation:
We are given that the function is defined as follows and
and we want to check the continuity in the interval [-4,5]. Note that this a piecewise function whose only critical point (that might be a candidate of a discontinuity) x=0 since at this point is where the function "changes" of definition. Note that 9-x and 9+12x are polynomials that are continous over all
. So F is continous in the intervals [-4,0) and (0,5]. To check if f(x) is continuous at 0, we must check that
(this is the definition of continuity at x=0)
Note that if x=0, then f(x) = 9-x. So, f(0)=9. On the same time, note that
. This result is because the function 9-x is continous at x=0, so the left-hand limit is equal to the value of the function at 0.
Note that when x>0, we have that f(x) = 9+12x. In this case, we have that
. As before, this result is because the function 9+12x is continous at x=0, so the right-hand limit is equal to the value of the function at 0.
Thus, , so by definition, f is continuous at x=0, hence continuous over the interval [-4,5].