please please help

Question

Y=a (x - p) ^2 + q

if 'p' is positive number and q & a is a negative number, then how many x intercepts will this quadratic function?

Answer 1

 

 Y=a (x - p) ^2 + q

To find  x intercepts  make y = 0 in the given quadratic function .

a (x - p) ^2 + q = 0 

a(x² -2px +p² ) + q = 0

ax² -2apx + p²a + q = 0

Compare the above obtained equation with quadratic equation Ax² + Bx + C = 0 .

A = a , B = -2ap  , C = p²a + q 

We can use the radical part of the quadratic formula to determine if the graph

of a quadratic function crosses the x-axis at two, one, or no points.

 

B² -4AC = (-2ap )²  - 4 (a) ( p²a + q)

B² -4AC = 4a²p²  - 4 (a) ( p²a + q)

B² -4AC = 4 [ a²p²  - (a) ( p²a + q) ]

B² -4AC = 4 [ a²p²  - p²a² -aq ]

B² -4AC = 4 [-aq]

B² -4AC = -4aq

Given a and q are negative so the product of aq is positive .

The term  -4aq is negative .

Hence B² -4AC < 0 .

So , the graph has zero x-intercepts because the number under

the radical sign is negative and therefore, x is not a real number.

 

Now let us consider a quadratic function with the given specifications that is  'p' is positive number and q & a is a negative number  to prove that the function doesn't have a x-intercepts  .

y= -4(x-2)^2 -1

 

Now graph the function:

 

So we can conclude that the quadratic function  doesn't have a x-intercepts  when 'p' is positive number and q & a is a negative number .