1 year ago

Can someone please help mee(20 points and i will give brainliest!!!)

1 answer:

3 0

The vertex is the lowest/highest point, depending on if the parabola opens upwards or downwards.

So because the vertex opens upwards, we are looking for the lowest point which is at (2,-3).

You might be interested in

kirill115 [55]

Vertex is (4,0)
Fhats all I know

6 0

3 years ago

levacccp [35]

Answer is C.36 mod 3

4 0

3 years ago

Nataly [62]

Answer:

y = 2/3x+3

Step-by-step explanation:

5 0

3 years ago

Tems11 [23]

The slope of the tangent line to $f$ at $x=-1$ is given by the derivative of $f$ at that point:

$f'(-1)=\displaystyle\lim_{x\to-1}\frac{f(x)-f(-1)}{x-(-1)}=\lim_{x\to-1}\frac{2x^2-2}{x+1}$

Factorize the numerator:

$2x^2-2=2(x^2-1)=2(x-1)(x+1)$

We have $x$ approaching -1; in particular, this means $x\neq-1$ , so that

$\dfrac{2x^2-2}{x+1}=\dfrac{2(x-1)(x+1)}{x+1}=2(x-1)$

Then

$f'(-1)=\displaystyle\lim_{x\to-1}\frac{2x^2-2}{x+1}=\lim_{x\to-1}2(x-1)=2(-1-1)=-4$

and the tangent line's equation is

$y-f(-1)=f'(-1)(x-(-1))\implies y-4x-2$

6 0

3 years ago

NeX [460]

Answer:

x=9/20 or x=0.45

Step-by-step explanation:

8 0

3 years ago