If the vertex of graph is (5,-4) and two point of the graph are (-7,0) & (-3,0)...what is the discriminant?

Question

anonymous · Answer

do you know what discriminant is at all?

anonymous · Answer

what does it show?

anonymous · Answer

we can recover the whole equation from this information. and you have a choice of methods. since the vertex is 
$$(5,-4)$$ it must be 
$$y=a(x-5)^2-4$$ and now you can replace x by 7, set the result = 0 and solve for a

anonymous · Answer

it only shows the vertex and two points of the parabola

anonymous · Answer

or you can say it must look like 
$$y=a(x-7)(x+3)$$ and solve that way

anonymous · Answer

use foil?

anonymous · Answer

also you can have 
$$y=ax^2+bx+c$$
and plug in all three points and solve system of equation

anonymous · Answer

actually you have too much information. i will solve one way, maybe tomas another

anonymous · Answer

vertex is (5,-4) so it must be 
$$y=a(x-5)^2-4$$ now you know that if x = 7, y = 0 so put 
$$0=a(7-5)^2-4$$
$$a(4)-4=0$$
$$4a=4$$
$$a=1$$ so your equation is 
$$y=(x-5)^2-4$$

anonymous · Answer

or since you know 
$$a=1$$ and it must be 
$$y=a(x-7)(x+3)$$ then it is 
$$y=(x-7)(x-3)$$ either way.

anonymous · Answer

notice in the first example i only used two points. this is all you need and this problem is ill posed. you can recover the equation from the vertex and one zero

anonymous · Answer

to find the disc, it would be 25+16

anonymous · Answer

so its 41?

anonymous · Answer

if you multiply out you get 
$$y=x^2-4 x-21$$

anonymous · Answer

then compute 
$$b^2-4ac=(-4)^2-4\times 1\times (-21)$$

anonymous · Answer

i get 100

anonymous · Answer

because the zeros are rational numbers, the discriminant must be a perfect square

anonymous · Answer

oh thats not one of the choices..
the choices are 

A. 16
B. 0
C. -13
D. 15

anonymous · Answer

ah that is because i made a mistake. i  thought you said the zeros were (7,0) and (-3,0)

anonymous · Answer

i missed the minus sign in front of the 7

anonymous · Answer

Oh

anonymous · Answer

so lets try again.

it must be 
$$y=a(x-5)^2-4$$ and you know if x = -3 you get 0

anonymous · Answer

ok i think it is just 
$$y=(x+3)(x+7)$$ and so it is 
$$y=x^2+10 x+21$$ and therefore 
$$b^2-4ac=16$$

anonymous · Answer

ok..thanks for showing the steps..