Vertex form is accomplished by completing the square on the quadratic. Do this by first setting the parabola equal to 0 then moving the constant over to the other side:

$x^2+6x=-9$

Now take half the linear term, square it, and add it to both sides. Our linear term is 6. Half of 6 is 3, and 3 squared is 9:

$x^2+6x+9=-9+9$

The reason we do this is to create a perfect square binomial on the left:

$(x+3)^2=0$ (obviously the 0 results from the addition of 9 and -9). Move the 0 back over to the other side and set the quadratic back equal to y:

$(x+3)^2+0=y$

This gives you a vertex of (-3, 0), which is a minimum value, since the parabola opens upwards.

6 0

3 years ago

1+2i/2+3i simplified

Aleonysh [2.5K]

8/13 + i/13 hope this helps

4 0

3 years ago

The vertices of triangle rst are r(–1,–2), s(2,–1), and t(4,2). If δjkl ∼ δrst and the length of kl is units , what is the lengt

NemiM [27]

Answer:

3.16 units

Step-by-step explanation:

It has been given that the triangles JKL and the triangle RST are congruent.

That implies that, the length of the side JK, KL, and JL is equivalent to the length of the sides RS, ST, and RT respectively.

Now, to find the length of JK we need to find the length of the side RS. The coordinates of the points R and S are $(-1,-2)$ and $(2,-1)$ .

The length of the side RS is equal to the distance between point R and S.

RS $=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

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

$=\sqrt{3^2+(-1+2)^2}$

$=\sqrt{9+1}=\sqrt{10}=3.16$

Now that we have the length of the side RS, and the triangles JKL and RST are congruent therefore, the length of the side JK is 3.16 units.

7 0

3 years ago

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