F ( x ) = ( x + 3 )² - 8 = x² + 6 x + 9 - 8 = x² + 6 x + 1
For the quadratic function:
Axis of symmetry is: x = -b/ 2 a, where: a = 1, b = 6 ( because it it in the form:
y = a x² + b x + c ).
Therefore: x = -6 / 2 = -3.
f ( - 3 ) = ( - 3 + 3 )² - 8 = 0 - 8 = - 8
And D = b² - 4 a c = 6² - 4 * 1 * 1 = 36 - 4 = 32 ( greater than 0 ). It means that there are 2 real solutions.
Answer: x = - 3 , vertex : ( - 3 , - 8 ), Number of real solutions : 2.
a 1 = - 3 , a 2 = - 3 , a 3 = - 8 , a 4 = 2.
Answer:
try this
Step-by-step explanation:
desmos graphing will give u answers
Answer:
and
.
Step-by-step explanation:
So I believe the problem is this:

where we are asked to find values for
and
such that the equation holds for any
in the equation's domain.
So I'm actually going to get rid of any domain restrictions by multiplying both sides by (x-3)(x+7).
In other words this will clear the fractions.


As you can see there was some cancellation.
I'm going to plug in -7 for x because x+7 becomes 0 then.




Divide both sides by -10:


Now we have:
with 
I notice that x-3 is 0 when x=3. So I'm going to replace x with 3.




Divide both sides by 10:


So
and
.
Answer:
a = √11 and b = 6
Step-by-step explanation:
Refer to attached picture for reference
for an right triangle with angle θ
we are given
cos θ = 5/6 = length of adjacent side / length of hypotenuse
hence
adjacent length = 5 units
hypotenuse length = 6 units
the missing side is the "opposite" length which we can find with the Pythagorean equation. in our case:
hypotenuse ² = adjacent ² + opposite² (rearrange)
opposite ² = hypotenuse ² - adjacent ²
opposite ² = 6² - 5²
opposite = √ (6²-5²) = √11
sin θ = opposite length / hypotenuse (substitute values above)
sin θ = √11 / 6
hence a = √11 and b = 6