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3 years ago

Identify the turning point of the function f(x)= x^2 -2x + 8 by writing its equation in vertex form . show work

1 answer:

7 0

The turning point of a parabola is its vertex.

$f(x)=x^2-2x+8\\
f(x)=x^2-2x+1+7\\
f(x)=(x-1)^2+7$
The vertex is $(1,7)$

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Commutative Property of Multiplication.

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Answer:

Step-by-step explanation:

f(x)=3(2^x)+6 so

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f(3)=3(8)+6

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Note: Enter your answer and show all the steps that you use to solve this problem in the space provided. Write and solve an equa

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Answer:

Steven is 15 and one half years old now

Step-by-step explanation:

Linear equations

Sometimes we need to know the value of a variable in a given equation and that variable is given as a polynomial of degree 1. Solving for that variable means isolating it and replacing the other known values

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$x-8\frac{1}{2}$

We know he was 7 years old then, so

$x-8\frac{1}{2}=7$

We need to know his actual age, so let's solve for x, adding $8\frac{1}{2}$ to each side of the equation

$x-8\frac{1}{2}+8\frac{1}{2}=7+8\frac{1}{2}$

Simplifying

$x=15\frac{1}{2}$

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3 years ago

Scientists want to estimate the mean weight gain of mice after they have been fed a special diet. From previous studies, it is k

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Answer:

One example is the equation:

2x + 6 = 4x/2 + 12/2

To solve this, we need to transpose like terms to the same side.

2x - 4x/2 = 12/2 - 6

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0 = 0

Since both sides are zero, it means that the equation has infinite number of solutions.

Step-by-step explanation:

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2 years ago

R (-3,1) and S (-1,3) are points on a circle. If RS is a diameter, find the equation of the circle.

ELEN [110]

Answer:

$\sf (x+2)^2+(y-2)^2=2$

Step-by-step explanation:

If RS is the diameter of the circle, then the midpoint of RS will be the center of the circle.

$\sf midpoint=\left(\dfrac{x_s-x_r}{2}+x_r,\dfrac{y_s-y_r}{2}+y_r \right)$

$\sf =\left(\dfrac{-1-(-3)}{2}+(-3),\dfrac{3-1}{2}+1 \right)$

$\sf =(-2, 2)$

Equation of a circle: $\sf (x-h)^2+(y-k)^2=r^2$

(where (h, k) is the center and r is the radius)

Substituting found center (-2, 2) into the equation of a circle:

$\sf \implies (x-(-2))^2+(y-2)^2=r^2$

$\sf \implies (x+2)^2+(y-2)^2=r^2$

To find $\sf r^2$ , simply substitute one of the points into the equation and solve:

$\sf \implies (-3+2)^2+(1-2)^2=r^2$

$\sf \implies 1+1=r^2$

$\sf \implies r^2=2$

Therefore, the equation of the circle is:

$\sf (x+2)^2+(y-2)^2=2$

4 0

2 years ago