Answer:
Graph shows the vertical asymptote at x nearing 4.
Explanation:
X must be more than 4 otherwise log(x-4) will be invalid;
Let us make a table of x & y
<span>x→→<span>log<span>(x−4)</span></span></span>
------- ----------------
<span>5→→0</span>
<span>9→→.7</span>
<span>14→→1</span>
<span>19→→1.18</span>
<span>24→→1.30</span>
Now we can plot the Graph with the value of x and y. On this Graph Vertical asymptote is at x nearing the value 4. (Graph Below)
Hope this helped :)
Answer:
The answer is A.) 9
Step-by-step explanation:
8.1845062
8.1845 << your answer.
the number 5 was in the hundred thousandths so to round it look at the next number and if it's less than 5, it stays the same but if it's more than 5 you round it up to the next number.
hope this helps, God bless!
The equation of the parabola could be written as y-k = a(x-h)^2, where (h,k) is the vertex. Thus, y-(-3) = a(x+4)^2, or y+3 = a(x+4)^2.
The coordinates of one x-intercept are (-11,0). Thus, y+3 = a(x+4)^2 becomes
0+3 = a(-11+4)^2, so that 3 = a(-7)^2, or 3 = 49a. Therefore, a = 3/49, and the equation of the parabola becomes
y+3 = (3/49)(x+4)^2.
To find the other x-intercept, let y = 0 and solve the resulting equation for x:
0+3 = (3/49)(x+4)^2, or (49/3)*2 = (x+4)^2
Taking the sqrt of both sides, plus or minus 49/3 = x+4.
plus 49/3 = x+4 results in 37/3 = x, whereas
minus 49/3 = x+4 results in x = -61/3. Unfortunatelyi, this disagrees with what we are told: that one x-intercept is x= -11, or (-11,0).
Trying again, using the quadratic equation y=ax^2 + bx + c,
we substitute the coordinates of the points (-4,-3) and (-11,0) and solve for {a, b, c}:
-3 = a(-4)^2 + b(-4) + c, or -3 = 16a - 4b + c
0 = a(-11)^2 - 11b + c, or 0 = 121a - 11b + c
If the vertex is at (-4,-3), then, because x= -b/(2a) also represents the x-coordinate of the vertex, -4 = b / (2a), or -8a = b, or
0 = 8a + b
Now we have 3 equations in 3 unknowns:
0 = 8a + 1b
-3 = 16a - 4b + c
0 = 121a - 11b + c
This system of 3 linear equations can be solved in various ways. I've used matrices, finding that a, b and c are all zero. This is wrong.
So, let's try again. Recall that x = -b / (2a) is the axis of symmetry, which in this case is x = -4. If one zero is at -11, this point is 7 units to the left of x = -4. The other zero is 7 units to the right of x = -4, that is, at x = 3.
Now we have 3 points on the parabola: (-11,0), (-4,-3) and (3,0).
This is sufficient info for us to determine {a,b,c} in y=ax^2+bx+c.
One by one we take these 3 points and subst. their coordinates into
y=ax^2+bx+c, obtaining 3 linear equations:
0=a(-11)^2 + b(-11) + 1c => 0 = 121a - 11b + 1c
-3 = a(-4)^2 +b(-4) + 1c => -3 = 16a - 4b + 1c
0 = a(3)^2 +b(3) + c => 0 = 9a +3b + 1c
Solving this system using matrices, I obtained a= 3/49, b= 24/49 and c= -99/49.
Then the equation of this parabola, based upon y = ax^2 + bx + c, is
y = (1/49)(3x^2 + 24x - 99) (answer)
Check: If x = -11, does y = 0?
(1/49)(3(-11)^2 + 24(-11) - 99 = (1/49)(3(121) - 11(24) - 99
= (1/49)(363 - 264 - 99) = (1/49)(0) YES!
y = (1/49)(3x^2 + 24x - 99) (answer)
No because then somewhere in the problem you could multiply once more. like if u had 234 as ur answer, but had 2 as a remainder, u would change the 4 to a 5 (don't know if tht math is right, but the main idea of it is right)
