Let f be the function defined as follows:
(attached)

if a = 2 and b = 3, is f continuous at x = 1? justify your answer.

find a relationship between a and b for which f is continuous at x = 1.
hint: a relationship between a and b just means an equation in a and b.

find a relationship between a and b so that f is continuous at x = 2.

use your equations from parts (ii) and (iii) to find the values of a and b so that f is continuous at both x = 1 and also at x = 2?

graph the piece function using the values of a and b that you have found. you may graph by hand or use your calculator to graph and copy and paste into the document.

a = -2; b = 4  

Step-by-step explanation:

I graphed your piecewise function in Figure 1.

The relation involving a and b is a parabola. It is the green segment in the middle of the graph. We must find values of a and b that make the parabola connect with the ends of the straight lines.

We could make the function continuous if the vertex were at (1, 2) and the x-intercept were at (2, 0).

The vertex form of the equation for a parabola is

y = a(x - h)² + k

where h and k are the coordinates of the vertex and a is a constant.

Data:

Vertex at (1, 2)

x-intercept at (2, 0)

Calculations:

1. Substitute the coordinates of the vertex into the equation

y = a(x - 1)² + 2

2. Substitute the coordinates of the x-intercept into the equation and solve for a

Thus, in the vertex form, a = -2, h = 1, k = 2

3. Determine the equation of the parabola in standard form

y = a(x - h)² + k = -2(x - 1)² + k = -2(x²- 2x + 1) + 2 = -2x² + 4x - 2 + 2

= -2x² + 4x

The values of a and b in the missing segment are -2 and + 4.

I graphed the missing parabola in Fig.2. It makes your piecewise function continuous.

< coa = 150 ( given)

as ba and bc are tangents

so oc and oa is perpendicular to cb and ab respectively because in circle normal passes through centre

so in quadrilateral oabc

< coa + < oab + < ocb + < abc = 360

150 +90+90 + < abc = 360

330 + < abc = 360

< abc = 360 -330 = 30

so < abc = 30

⭐dhruv⭐

69.64%

step-by-step explanation: