Which function has x-intercepts of 3 and 8 and contains the point (–1, –2)?

1 answer:

6 0

This is how you do it.

y = a(x - 3)(x - 8)
-2 = a(-1 - 3)(-1 - 8) 
-2 = 36a 
- 1 / 18 = a 
y = ( - 1 / 18)(x^2 - 11x + 24) 
y = (- 1 / 18)x^2 + (11 / 18)x - (3 / 2)

If this doesn't explain it enough, please, ask questions.

You might be interested in

On a piece of paper graph y=x+2 then determine which answer choice matches the graph you drew use the graph to find the zero of

kolezko [41]

Answer:

Step-by-step explanation:

When x=2 it reaches zero on the graph and also has the correct transformation

4 0

3 years ago

Read 2 more answers

A fisherman can row upstream at 4mph and downstream at 6mph. He started rowing upstream until he got tired and then started down

ikadub [295]

Total time = upstream time + downstream time
he rowed the same distance both ways.
upstream time = D/4
downstream time = D/6
total time = D/4 + D/6 = 3D/12 + 2D/12 = 5D/12 = 3 hours
D = 3*12/5 = 36/5 or 7.2 miles
where D is the one-way distance. The total distance he rowed is 2*D = 72/5 or 14.4 miles

6 0

3 years ago

Read 3 more answers

19. Hillary buys 3 pounds of Gala apples and some Granny Smith apples. Both kinds of apples cost $4.50 per pound. The total cos

GarryVolchara [31]

Answer:

2.5 pounds of Granny Smith apples.

Step-by-step explanation:

7 0

3 years ago

Read 2 more answers

Use the following number line to determine if the expressions are true or false.

Usimov [2.4K]

True cus from point a and b is very low so true slime

6 0

3 years ago

Find the max and min values of f(x,y,z)=x+y-z on the sphere x^2+y^2+z^2=81

Anton [14]

Using Lagrange multipliers, we have the Lagrangian

$L(x,y,z,\lambda)=x+y-z+\lambda(x^2+y^2+z^2-81)$

with partial derivatives (set equal to 0)

$L_x=1+2\lambda x=0\implies x=-\dfrac1{2\lambda}$
$L_y=1+2\lambda y=0\implies y=-\dfrac1{2\lambda}$
$L_z=-1+2\lambda z=0\implies z=\dfrac1{2\lambda}$
$L_\lambda=x^2+y^2+z^2-81=0\implies x^2+y^2+z^2=81$

Substituting the first three equations into the fourth allows us to solve for $\lambda$ :

$x^2+y^2+z^2=\dfrac1{4\lambda^2}+\dfrac1{4\lambda^2}+\dfrac1{4\lambda^2}=81\implies\lambda=\pm\dfrac1{6\sqrt3}$

For each possible value of $\lambda$ , we get two corresponding critical points at $(\mp3\sqrt3,\mp3\sqrt3,\pm3\sqrt3)$ .

At these points, respectively, we get a maximum value of $f(3\sqrt3,3\sqrt3,-3\sqrt3)=9\sqrt3$ and a minimum value of $f(-3\sqrt3,-3\sqrt3,3\sqrt3)=-9\sqrt3$ .

5 0

3 years ago