Question

Q.NO 4: Find a function y=ax^2+bx+c
whose graph has an x intercept of 1, a y-intercept of -2
and a tangent line with a slope of - 1 at the y-intercept?​

Answer 1

Answer:

solved

Step-by-step explanation:

y = ax²+bx+c

y = 0 at x = 1

a+b+c = 0

x = 0 , at y = -2 , c = -2 → a+b = 2

slope of tangent = dy/dx = 2ax+b = b = -1 → a = 3

y = 3x² - x - 2

Answer 2

Answer:

The required function for which the given information is satisfied, is found to be: $y=3x^2-x-2$

Step-by-step explanation:

The given equation of the function is as follows:

$y=ax^2+bx+c$                     ...(i)

Now, we are given a x-intercept at 1, means at x=1, the function crosses the x-axis, i.e. the point on the graph will be: (1,0)

Substituting this coordinate pair in the equation (i), we get:

$0=a(1)^2+b(1)+c$

Simplifying it, we get:

$a+b+c=0$                           ...(ii)

Similarly, we are given a y-intercept at -2, means at y=-2, the function crosses the y-axis, i.e. the point on the graph will be: (0,-2)

Substituting this coordinate pair in the equation (i), we get:

$-2=a(0)^2+b(0)+c$

Simplifying it, we get:

$c=-2$                                    ...(iii)

Now, we are given a tangent of the equation (i) as -1. The expression for the same becomes:

$\frac{d}{dx}(ax^2+bx+c)=-1$

Simplifying it, we get:

$2ax+b=-1$        

Substituting (0,-2) in this, we get:

$b=-1$                                   ...(iv)

Substituting (iii) and (iv) in (ii), we get:

$a-1-2=0$

$a=3$

Now, we have got all the coefficients of equation (i), so (i) becomes:

$y=3x^2-x-2$

which is the required answer.