SOLUTION: f(x)=x^2-4x-5 i need help finding y intercept, x intercept, vertex, and axis of symmetry i am not grasping how to get the answer even after looking over other examples..

Question 760887: f(x)=x^2-4x-5 i need help finding y intercept, x intercept, vertex, and axis of symmetry i am not grasping how to get the answer even after looking over other examples..
Found 3 solutions by Cromlix, solver91311, lwsshak3:
Answer by Cromlix(4381)   (Show Source): You can put this solution on YOUR website!
f(x) = x^2 - 4x - 5
Factorise
(x - 5)(x + 1) = 0
x - 5 = 0
x = 5
x + 1 = 0
x = -1
This where the parabola cuts the x axis (-1,0) and (5,0)
f(x) = x^2 - 4x - 5
To find where it cuts the y axis make x = 0
f(0) = 0^2 -4(0) - 5
So parabola cuts y axis at (0, -5)
To find the vertex.
The distance between x = -1 and x = 5
(where it cuts the x axis) is 6 units
Half of 6 = 3
So either add 3 to -1 = 2
or subtract 3 from 5 = 2
Now this shows that the axis of symmetry is
x = 2
Now put x =2 into the equation
f(x) = x^2 - 4x - 5
= (2)^2 - 4(2) - 5
= 4 - 8 - 5
= -9
This is the y coordinate of the vertex
(2, -9)
I hope this helps you.
:-)
Answer by solver91311(24713)   (Show Source): You can put this solution on YOUR website!

Given

The -coordinate of the -intercept is given by:

Hence the -intercept is the point

The -coordinates of the -intercepts, if they exist are the real number results of the quadratic formula.

Note: 2 real and distinct roots are guaranteed when the signs on coefficients and are opposite.

The -intercepts are then and

The -coordinate of the vertex is

The -coordinate of the vertex is:

The axis of symmetry is the vertical line that passes through the vertex.

All you have to do is plug in the numbers for your particular , , and (remembering the binding rules for the minus signs) into the above and do the indicated arithmetic.

John

Egw to Beta kai to Sigma
My calculator said it, I believe it, that settles it

Answer by lwsshak3(11628)   (Show Source): You can put this solution on YOUR website!
f(x)=x^2-4x-5 i need help finding y intercept, x intercept, vertex, and axis of symmetry
***
y=x^2-4x-5
To find y-intercept, set x=0, then solve for y:
y=0-0-5
y-intercept=-5
..
To find x-intercepts, set y=0 then solve for x:
0=x^2-4x-5
factor:
(x-5)(x+1)=0
x=5
and
x=-1
x-intercepts are 5 and -1
..
To find the vertex:
Complete the square to write equation in standard form:
y=x^2-4x-5
y=(x^2-4x+4)-4-5
y=(x-2)^2-9
This is an equation of a parabola that opens upward.
Its standard (vertex) form:y=(x-h)^2+k, (h,k)=(x,y) coordinates of the vertex
vertex: (2,-9)
axis of symmetry: x=2
see graph below as a visual check:

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