Question

PLEASE HELP i AM LOST AT THIS

Find the vertex of parabola

f(x)=x^2-10-3

Answer 1

completing the square might be useful; or if you know the vertex part of the quadratic formula

Answer 2

given: ax^2 + bx + c

the x value of the vertex has the form: -b/(2a)

Answer 3

and i believe the y part would amount to: c - (b/2a)^2

Answer 4

the problem is x^2-10x-3

Answer 5

can you define the abc parts?

Answer 6

yes this is what I have so far on this problem

x=- -10/2*1 right

Answer 7

x = --10/2*1 is good to me, simplifies to 5

Answer 8

lets make this simpler for y, i have a notion that once you find x; then y= c - x^2

Answer 9

so, y= -3 - (5)^2

Answer 10

so it would not be (5)^2-10*5-3 then that is where I get lost

Answer 11

it would be, but thats a rather brute way to go about it; it gets the same results tho

if i take my idea from the method of completing the square:

ax^2 + bx + c

x^2 + b/a x + c/a

x^2 + b/a x +(b/2a)^2 -(b/2a)^2 + c/a

(x^2 + b/a x +(b/2a)^2) -(b/2a)^2 + c/a

(x - -(b/2a))^2 + ( c/a - (b/2a)^2 )
^^ ^^^^^^^^^^^^
X vertex Y vertex

so, x=-b/2a ; and y= c/a - x^2 .... i was close, but since a=1 it was immaterial

Answer 12

so the vertex of the fiven parabola is (5,??) what is the last number?

Answer 13

given sorry

Answer 14

y= -3 - (5)^2 is what i keep posting :)

Answer 15

if we go the brute method

(5)^2: 25
-10*5:-50
-3: -3
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y : 25-53

im really not sure where the confusion is at