Functions and Relations.(If f(x + 1) = x² + 4x - 5, find f(x - 2)

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[yosualeonardo]

Nice explanation!

I don't know if my way is correct but here is what I do:
f(x+1) = x²+4x-5
f(x+1) = (x+5)(x-1)
f(x+1) = (x+1+4)(x+1-2)
f(x-2) = (x-2+4)(x-2-2)
f(x-2) = (x+2)(x-4)
f(x-2) = x²-2x-8

[myelectronicsworld]

You could have set x=u-3 from the very beginning and you get the same result.

[mathphobia]

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[samkay]

Very nice explanation.
Thank you so much...

[aitheria]

Good video with good explanation. Thanks man i love it ❤

[eswyatt]

So much of math is procedure (Fourier Series, solving canonical PDEs, . . .). What's happening here is really thinking!!!

[赖皮球]

f(x + 1) = x² + 4x - 5, f(x + 1) = （x+1）² + 2(x +1)- 8, , , so f(x-2)=（x-2）² + 2(x -2)- 8=x² -4x+4+2x-4-8=x² -2x-8

[hamoudiayoub]

f(x-2) = f((x-3)+1)
= (x-3)² + 4(x-3) - 5
= x² - 2x - 8

[MrLidless]

f(x) = ax^2 + bx + c
f(x + 1) = ax^2 + (2a + b)x + (a + b + c) = x^2 + 4x - 5
a = 1, b = 2, c = -8
f(x) = (x + 4)(x - 2)
f(x - 2) = (x + 2)(x - 4)

[NurHadi-qfkl]

.x-1=x+1-3
.f=(x-3)^2+4(x-3)-5=x^2-2x-3

[samuelweir]

I would have just noted that if you make the substitution x->x-3 on both sides of f(x+1)=x^2+4x-5, then f(x+1) becomes f(x-2).