Guided Practice Write the equation of the parabola in vertex form. The vertex is (−3, 3) and the parabola passes through the point (4, 150). Create a rough sketch of the graph to use as a visual aid.

A.
y = 3(x + 3)^2 + 3

B.
y = 4(x − 3)^2 + 3

C.
y = 50(x + 3)^2 + 4

D.
y = 3(x − 3)^2 + 3

A.

y = 3(x + 3)^2 + 3

Step-by-step explanation:

The vertex form of the equation is y=a(x-h)2 + k  

You know that the vertex (h,k)  is (-3,3) and a point on the parabola is (4,150). So what you’re needed ing to find is a bc all of the other letters can be replaced into the equation.  

First I would put the equation into vertex form using the vertex given in the problem. : y= (x - (-3))2 + (3). Then I would use the ordered pair (4,150) and replace the x and y with those values

150 = a(4 + 3)2 + 3  

Now simplify

150 = a (7)2 + 3  

150 = 49a + 3      Important: please be sure to square.

147 = 49a

3= a  

Next replace a into the formula now leaving x and y alone.

y = 3 (x+3)2 + 3   [Vertex form]

I hope this helps. Have a great day!

m∠cdf = 43° , m∠gdf = 114°, m∠gde = 23°, m∠edh = 23° , m∠cdh = 157°

step-by-step explanation:

since, it is given that de bisects the angle gdh ⇒ de divides the angle gdh into two equal angles.

⇒ m∠gde = m∠edh

⇒ 8·x   - 1 = 6·x + 5

⇒ x = 3

so, m∠gde = 8·x - 1 = 23°

and m∠edh = 23°

now, it is given ∠cde is an straight angle so, m∠cde = 180°

⇒ m∠cdf + m∠fdg + m∠gde = 180°

⇒ 43° + m∠fdg + 23° = 180°

⇒ m∠fdg = 114°

again, m∠cde = m∠cdh + m∠edh = 180°

m∠cdh + 23° = 180°

⇒ m∠cdh = 157°

the weight of the object inside the box!