Answer:
Any expression of the form 
Answer:
y = -(x + 5)² + 4
Step-by-step explanation:
The roots are -3 and -7, so:
y = a (x + 3) (x + 7)
Distribute and complete the square:
y = a (x² + 10x + 21)
y = a (x² + 10x + 25 − 4)
y = a (x² + 10x + 25) − 4a
y = a (x + 5)² − 4a
The vertex is (-5, 4), so a = -1.
y = -(x + 5)² + 4
Answer:
<h2>3) -82.</h2><h2>4)-70.</h2><h2>5) -15.</h2><h2>6)-14.</h2><h2>7)5</h2><h2>8)15</h2><h2>9)16</h2><h2>10 )16</h2><h2>11)64</h2><h2>12)21.</h2><h2>13)1500 in the middle (not above not below)</h2><h2>14)450 m</h2>
Answer:
Step-by-step explanation:
We are asked the equation of the graph which passes through the points at (0,5) and (-3,-4).
Now, we know that the equation of a straight line passing through the points (
) and (
) is given by
.
So, in our case () ≡ (0,5) and () ≡ (-3,-4)
Therefore, the equation of the straight line will be
⇒ 
⇒ (Answer)
Answer:
See explanation and hopefully it answers your question.
Basically because the expression has a hole at x=3.
Step-by-step explanation:
Let h(x)=( x^2-k ) / ( hx-15 )
This function, h, has a hole in the curve at hx-15=0 if it also makes the numerator 0 for the same x value.
Solving for x in that equation:
Adding 15 on both sides:
hx=15
Dividing both sides by h:
x=15/h
For it be a hole, you also must have the numerator is zero at x=15/h.
x^2-k=0 at x=15/h gives:
(15/h)^2-k=0
225/h^2-k=0
k=225/h^2
So if we wanted to evaluate the following limit:
Lim x->15/h ( x^2-k ) / ( hx-15 )
Or
Lim x->15/h ( x^2-(225/h^2) ) / ( hx-15 ) you couldn't use direct substitution because of the hole at x=15/h.
We were ask to evaluate
Lim x->3 ( x^2-k ) / ( hx-15 )
Comparing the two limits h=5 and k=225/h^2=225/25=9.
