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)
IJKL is a rectangle, so opposite sides have the same length:
• IJ = KL ⇒ 6y - 6 = 2x + 20
• JK = IL ⇒ 3x + 21 = 6y
Substitute the second equation into the first and solve for x :
6y - 6 = 2x + 20
(3x + 21) - 6 = 2x + 20
3x + 15 = 2x + 20
x = 5
Solve for y :
3x + 21 = 6y
15 + 21 = 6y
36 = 6y
y = 6
Answer:
The exponential function that passes through (2,36) is:
.
Step-by-step explanation:
We are asked to find which function passes through the point (2,36).
i.e. we will put the input value '2' in the following given functions and check which gives the output value as '36'.
1)
now we put x=2.
hence option 1 is correct.
2)
Now we put x=2.
Hence, option 2 is incorrect.
3)
Now we put x=2
Hence, option 3 is incorrect.
4)
Now we put x=2.
Hence, option 4 is incorrect.
Hence, option 1) is correct.
i.e. The exponential function that passes through (2,36) is:
R parallel to s
because <4 =<5 alternate
