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Sure, let's solve the equation [tex]\( y = x^2 - 2x + 1 \)[/tex] step-by-step.
1. Identify the given equation:
The equation to solve is [tex]\( y = x^2 - 2x + 1 \)[/tex].
2. Rewrite the equation:
Our equation in expanded form is already in the standard quadratic form [tex]\( ax^2 + bx + c \)[/tex], where [tex]\( a = 1 \)[/tex], [tex]\( b = -2 \)[/tex], and [tex]\( c = 1 \)[/tex].
3. Factor the quadratic equation:
Notice that [tex]\( y = x^2 - 2x + 1 \)[/tex] can be factored as:
[tex]\[ y = (x - 1)^2 \][/tex]
4. Confirm the factoring:
To ensure the factorization is correct, expand [tex]\( (x - 1)^2 \)[/tex]:
[tex]\[ (x - 1)^2 = (x - 1)(x - 1) = x^2 - x - x + 1 = x^2 - 2x + 1 \][/tex]
This matches the original quadratic equation, so the factorization is correct.
5. Interpret the factorized form:
The equation [tex]\( y = (x - 1)^2 \)[/tex] indicates that [tex]\( y \)[/tex] is a perfect square of the expression [tex]\( x - 1 \)[/tex]. This form tells us that [tex]\( y \)[/tex] achieves its minimum value when [tex]\( x - 1 = 0 \)[/tex], i.e., when [tex]\( x = 1 \)[/tex].
6. Find the vertex of the parabola:
The vertex form of a parabola [tex]\( y = a(x - h)^2 + k \)[/tex] shows that the vertex of the parabola [tex]\( y = (x - 1)^2 \)[/tex] is at the point [tex]\( (h, k) \)[/tex]. Here, [tex]\( h = 1 \)[/tex] and [tex]\( k = 0 \)[/tex].
7. Analyze the function:
Since [tex]\( (x - 1)^2 \)[/tex] is always non-negative, [tex]\( y \geq 0 \)[/tex]. The minimum value of [tex]\( y \)[/tex] is [tex]\( 0 \)[/tex] and it occurs at [tex]\( x = 1 \)[/tex].
8. Final interpretation:
[tex]\[ y = x^2 - 2x + 1 = (x - 1)^2 \][/tex]
This tells us that the quadratic function [tex]\( y = x^2 - 2x + 1 \)[/tex] represents a parabola that opens upwards with its vertex at [tex]\( (x, y) = (1, 0) \)[/tex].
Thus, the detailed solution for the equation [tex]\( y = x^2 - 2x + 1 \)[/tex] is [tex]\( y = (x - 1)^2 \)[/tex], which indicates that the quadratic function can be represented as a parabola with a vertex at [tex]\( (1, 0) \)[/tex], achieving a minimum value of [tex]\( 0 \)[/tex] when [tex]\( x = 1 \)[/tex].
1. Identify the given equation:
The equation to solve is [tex]\( y = x^2 - 2x + 1 \)[/tex].
2. Rewrite the equation:
Our equation in expanded form is already in the standard quadratic form [tex]\( ax^2 + bx + c \)[/tex], where [tex]\( a = 1 \)[/tex], [tex]\( b = -2 \)[/tex], and [tex]\( c = 1 \)[/tex].
3. Factor the quadratic equation:
Notice that [tex]\( y = x^2 - 2x + 1 \)[/tex] can be factored as:
[tex]\[ y = (x - 1)^2 \][/tex]
4. Confirm the factoring:
To ensure the factorization is correct, expand [tex]\( (x - 1)^2 \)[/tex]:
[tex]\[ (x - 1)^2 = (x - 1)(x - 1) = x^2 - x - x + 1 = x^2 - 2x + 1 \][/tex]
This matches the original quadratic equation, so the factorization is correct.
5. Interpret the factorized form:
The equation [tex]\( y = (x - 1)^2 \)[/tex] indicates that [tex]\( y \)[/tex] is a perfect square of the expression [tex]\( x - 1 \)[/tex]. This form tells us that [tex]\( y \)[/tex] achieves its minimum value when [tex]\( x - 1 = 0 \)[/tex], i.e., when [tex]\( x = 1 \)[/tex].
6. Find the vertex of the parabola:
The vertex form of a parabola [tex]\( y = a(x - h)^2 + k \)[/tex] shows that the vertex of the parabola [tex]\( y = (x - 1)^2 \)[/tex] is at the point [tex]\( (h, k) \)[/tex]. Here, [tex]\( h = 1 \)[/tex] and [tex]\( k = 0 \)[/tex].
7. Analyze the function:
Since [tex]\( (x - 1)^2 \)[/tex] is always non-negative, [tex]\( y \geq 0 \)[/tex]. The minimum value of [tex]\( y \)[/tex] is [tex]\( 0 \)[/tex] and it occurs at [tex]\( x = 1 \)[/tex].
8. Final interpretation:
[tex]\[ y = x^2 - 2x + 1 = (x - 1)^2 \][/tex]
This tells us that the quadratic function [tex]\( y = x^2 - 2x + 1 \)[/tex] represents a parabola that opens upwards with its vertex at [tex]\( (x, y) = (1, 0) \)[/tex].
Thus, the detailed solution for the equation [tex]\( y = x^2 - 2x + 1 \)[/tex] is [tex]\( y = (x - 1)^2 \)[/tex], which indicates that the quadratic function can be represented as a parabola with a vertex at [tex]\( (1, 0) \)[/tex], achieving a minimum value of [tex]\( 0 \)[/tex] when [tex]\( x = 1 \)[/tex].
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