Solve. -317 - 221 - (-354) A. 892 B. -892 C. 184 D. -184

If the vertex of a parabola is (-4, 6) and another point on the curve is (-3, 14), what is the coefficient of the squared expres

Luda [366]

$\bf \qquad \textit{parabola vertex form}\\\\\begin{array}{llll}
\boxed{y=a(x-{{ h}})^2+{{ k}}}\\\\
x=a(y-{{ k}})^2+{{ h}}
\end{array}\qquad\qquad vertex\ ({{ h}},{{ k}})\\\\
-------------------------------\\\\
vertex\ (-4,6)\qquad 
\begin{cases}
h=-4\\
k=6
\end{cases}\implies y=a(x-(-4))^2+6
\\\\\\
y=a(x+4)^2+6
\\\\\\
\textit{we also know that }
\begin{cases}
x=-3\\
y=14
\end{cases}\implies 14=a(-3+4)^2+6$

solve for "a"

8 0

3 years ago

How many solutions are there to the equation below?

romanna [79]

Its Infinitely many solutions then the answer is A

5 0

3 years ago

How do you solve this??? I need it fast please help with steps. (x+3)(x+1)^2(x−4)>0

scZoUnD [109]

Answer:

x < -3 or x > 4

Step-by-step explanation:

The product of the factors is 4th-degree, and the leading coefficient is 1 (positive).

The factors tell you the following about the zeros:

x = -3 . . . sign change from + to -

x = -1 . . . graph touches, but no sign change. - on either side of x=-1

x = 4 . . . sign change from - to +

__

The function will be positive for x < -3 and for x > +4.

_____

Additional comments

The value of the function is zero when any of its factors is zero. The value of a factor changes sign when the value of x changes from one side of the 0 to the other. For example, here, we have x+1 as a factor, so x=-1 is a zero. When x = -0.9, the factor is positive (-0.9 +1 = 0.1). When x = -1.1, the factor is negative (-1.1 +1 = -0.1). If this factor were to the first power, it would cause the value of the function to change sign at x=-1.

However, the factor (x+1) has an even power: (x+1)^2. That means when the factor is negative, its square is positive, and when the factor is positive, its square is positive. In short, the factor (x+1)^2 causes the function to be zero at x=-1, but does not make the function change sign there.

The product of all of these factors will result in a polynomial with x^4 as the highest-degree term. That means the function is of even degree. The leading coefficient is 1, so is positive, and the function will generally have a U-shape. The left-most (odd-degree) zero will be where the function changes sign from positive to negative. The right-most (odd-degree) zero will be where the function changes sign from negative to positive. Those zeros are x=-3 and x=+4, respectively. There are no places between these where the function changes sign, so these zeros are the ends of the regions where the function is positive (>0).

3 0

2 years ago

How do I graph that

nika2105 [10]

Start by finding the vertex, which is the point the parabola turns. The x value is -b/2a, which you have in your notes here. b = -8 and a = 2 and therefore the x value of the vertex is 2. Plug that in to find the y value, which is -2.

Start by plotting this point and the y intercept, which is 6. The parabola will be a U shape, which goes upwards starting at the vertex. If you draw that using the side with the y intercept, you can just mirror it on the other side of the vertex.

7 0

3 years ago

What is the area of the shaded region to the nearest tenth?

Aleks04 [339]

Answer:

15.1

Step-by-step explanation:

We know that the area of the circle is $\pi r^{2}$ , but we are looking for the portion of the circle, this means we multiply it by the portion of the circle that is shaded.

Area of shaded region = $\pi r^{2} (\frac{x}{360} )$

where r = 4 and x = 108, the 360 comes from the fact that the total degrees around the circle is 360, so dividing the total amount of degrees by the angle will give us the portion we are looking for.

Area of shaded region = $\frac{108}{360} \pi 4^{2}$ = 15.1

Another way to think about it is that the area of the total circle is $\pi r^{2}$ , multiplying it by a fraction, a number less than 1, would give us a smaller value.

Hope this helps.

5 0

3 years ago