Draw a sketch graph of p (x) =ax^2+bx+c from the following information

Question

1) b/2a = - 1 2) the range of the function is y < or equal to 3 3) the roots of p(x) differ by 4 Also draw the axis of symmetry and indicate the intercepts on the x-axis

Answer 1

Let me try,

P(x) = ax^2 + bx +c

So the vertex is at P'(x)=0

P'(x) = 2ax + b = 0

=> x = -b/2a

but given, b/2a = -1

so, x = 1

Therefore the axis of symmetry is at x=1

And let x-intercept be x1 and x2

So, (x1 + x2)/2 = x

=> x1 + x2 = 2*x = 2*1 = 2

=> x1 + x2 = 1   ------------------------ (1)

Also,

Given, x1 - x2 = 4  ------------------(2)

On solving eq (1) and eq(2) we get

x1=3 and x2= -1

so the equation is y = x^2 - (x1+x2)x + x1*x2

y = x^2 - 2x - 3

So, a = 1, b= -2  and c = -3

So, y position of vertex is given by  -(b^2 - 4ac)/4a

=> y(vertex) = -(4+12)/4 = -4

So the coordinate of vertex is (1,-4)

Also y-intercept

y = 0^2- 2*0 -3 = -3

So,

Complete Solution:

x intercepts = -1 and 3

y intercept = -3

axis of symmetry: x=1

vertex position: (1,-4)

graph for y<=3 is in link below:

https://ibb.co/Js09K31

Comments

Typing error:

=> x1 + x2 = 1   ------------------------ (1)

should be

=> x1 + x2 = 2   ------------------------ (1)

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Thank you for helping

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Thank You