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Sagot :
To analyze the properties of the graph described by the equation [tex]\( y = x^2 + x + 1 \)[/tex], we will consider several aspects: symmetry, y-intercept, and intersection with the x-axis.
1. Symmetry about the [tex]\( y \)[/tex]-axis:
To determine whether the graph is symmetric about the [tex]\( y \)[/tex]-axis, we need to check if [tex]\( y(x) \)[/tex] satisfies [tex]\( y(x) = y(-x) \)[/tex]. Replacing [tex]\( x \)[/tex] with [tex]\( -x \)[/tex]:
[tex]\[ y(-x) = (-x)^2 + (-x) + 1 = x^2 - x + 1 \][/tex]
Since [tex]\( y(-x) \neq y(x) \)[/tex] (because [tex]\( x^2 - x + 1 \neq x^2 + x + 1 \)[/tex]), the graph is not symmetric about the y-axis. Therefore, we can rule out option A.
2. Y-intercept:
The y-intercept is found by setting [tex]\( x = 0 \)[/tex] in the equation:
[tex]\[ y = 0^2 + 0 + 1 = 1 \][/tex]
The y-intercept is 1, which is positive. Thus, the graph does not have a negative y-intercept. Hence, option B is eliminated.
3. Intersection with the x-axis:
To find where the graph intersects the x-axis, we solve the equation [tex]\( x^2 + x + 1 = 0 \)[/tex]:
The discriminant [tex]\( \Delta \)[/tex] of a quadratic equation [tex]\( ax^2 + bx + c \)[/tex] is given by:
[tex]\[ \Delta = b^2 - 4ac \][/tex]
Substituting [tex]\( a = 1 \)[/tex], [tex]\( b = 1 \)[/tex], and [tex]\( c = 1 \)[/tex]:
[tex]\[ \Delta = 1^2 - 4(1)(1) = 1 - 4 = -3 \][/tex]
Since the discriminant is negative ([tex]\( \Delta < 0 \)[/tex]), the equation [tex]\( x^2 + x + 1 = 0 \)[/tex] has no real roots. This means the graph of the quadratic does not intersect the x-axis. Consequently, the graph does not have two points of intersection (option C) and is not tangent to the x-axis (option E).
Given this analysis, we conclude:
- The graph is not symmetric about the y-axis.
- The graph does not have a negative y-intercept.
- The graph does not intersect the x-axis.
Thus, the correct answer is:
D) The graph does not intersect the [tex]\( x \)[/tex]-axis.
1. Symmetry about the [tex]\( y \)[/tex]-axis:
To determine whether the graph is symmetric about the [tex]\( y \)[/tex]-axis, we need to check if [tex]\( y(x) \)[/tex] satisfies [tex]\( y(x) = y(-x) \)[/tex]. Replacing [tex]\( x \)[/tex] with [tex]\( -x \)[/tex]:
[tex]\[ y(-x) = (-x)^2 + (-x) + 1 = x^2 - x + 1 \][/tex]
Since [tex]\( y(-x) \neq y(x) \)[/tex] (because [tex]\( x^2 - x + 1 \neq x^2 + x + 1 \)[/tex]), the graph is not symmetric about the y-axis. Therefore, we can rule out option A.
2. Y-intercept:
The y-intercept is found by setting [tex]\( x = 0 \)[/tex] in the equation:
[tex]\[ y = 0^2 + 0 + 1 = 1 \][/tex]
The y-intercept is 1, which is positive. Thus, the graph does not have a negative y-intercept. Hence, option B is eliminated.
3. Intersection with the x-axis:
To find where the graph intersects the x-axis, we solve the equation [tex]\( x^2 + x + 1 = 0 \)[/tex]:
The discriminant [tex]\( \Delta \)[/tex] of a quadratic equation [tex]\( ax^2 + bx + c \)[/tex] is given by:
[tex]\[ \Delta = b^2 - 4ac \][/tex]
Substituting [tex]\( a = 1 \)[/tex], [tex]\( b = 1 \)[/tex], and [tex]\( c = 1 \)[/tex]:
[tex]\[ \Delta = 1^2 - 4(1)(1) = 1 - 4 = -3 \][/tex]
Since the discriminant is negative ([tex]\( \Delta < 0 \)[/tex]), the equation [tex]\( x^2 + x + 1 = 0 \)[/tex] has no real roots. This means the graph of the quadratic does not intersect the x-axis. Consequently, the graph does not have two points of intersection (option C) and is not tangent to the x-axis (option E).
Given this analysis, we conclude:
- The graph is not symmetric about the y-axis.
- The graph does not have a negative y-intercept.
- The graph does not intersect the x-axis.
Thus, the correct answer is:
D) The graph does not intersect the [tex]\( x \)[/tex]-axis.
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