The vertex of the function is at x = -3/2. We cannot find the y value with an incomplete equation.
To find the x value of a vertex, we use the equation -b/2a, in which a is equal to the coefficient of x^2 (in this case 1) and b is equal to the coefficient of x (in this case 3).
-b/2a
-3/2(1)
-3/2
This gives us that value. To find the y value, you would plug this into the equation. Since this equation is not complete, you cannot do that yet.
IF you mean

, then the following would be a solution:
First add 16 to both sides of the equation:

Now divide both sides of the equation by 3:

Take the square root of both sides to solve for "X", or simply solve by inspection: "what number(s) can be squared to get 49?"
Answer: -7 and 7 So the smallest solution would be "-7"
Rounded to one decimal place means to round to the nearest tenth (i.e. to the nearest 0.1)
4.25 (rounded to the nearest tenth) = <u><em>4.3</em></u>
<span>5 n C r 3 = 10 </span><span>がんばってね</span>
Problem 1
Draw a straight line and plot P anywhere on it. Use the compass to trace out a faint circle of radius 8 cm with center P. This circle crosses the previous line at point Q.
Repeat these steps to set up another circle centered at Q and keep the radius the same. The two circles cross at two locations. Let's mark one of those locations point X. From here, we could connect points X, P, Q to form an equilateral triangle. However, we only want the 60 degree angle from it.
With P as the center, draw another circle with radius 7.5 cm. This circle will cross the ray PX at location R.
Refer to the diagram below.
=====================================================
Problem 2
I'm not sure why your teacher wants you to use a compass and straightedge to construct an 80 degree angle. Such a task is not possible. The proof is lengthy but look up the term "constructible angles" and you'll find that only angles of the form 3n are possible to make with compass/straight edge.
In other words, you can only do multiples of 3. Unfortunately 80 is not a multiple of 3. I used GeoGebra to create the image below, as well as problem 1.