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1. 在平面直角坐标系 xOyx O y 中，已知抛物线 y=x2−4x+cy=x^{2}-4 x+c. (1)抛物线经过点 (0,y1)\left(0, y_{1}\right) ...
Mar 9, 2024
1. 在平面直角坐标系 xOyx O y 中，已知抛物线 y=x2−4x+cy=x^{2}-4 x+c. (1)抛物线经过点 (0,y1)\left(0, y_{1}\right) 和 (3,y2)\left(3, y_{2}\right) ，请比较 y1y_{1} 和 y2y_{2} 的大小; (2)若抛物线与 xx 轴的一个交点的坐标为 (1,0)(1,0). (1)求抛物线与 xx 轴的另一个交点的坐标; (2)抛物线上两点 A(x0,m)，B(x0+3,n)A\left(x_{0}, m\right) ， B\left(x_{0}+3, n\right) ，满足 m n<0 ，求 x0x_{0} 的取值范围.
Solution by Steps
step 1
To compare y1 y_1 and y2 y_2 , substitute the given points into the equation y=x2−4x+c y = x^2 - 4x + c
step 2
For point (0,y1) (0, y_1) , substitute x=0 x = 0 to get y1=c y_1 = c
step 3
For point (3,y2) (3, y_2) , substitute x=3 x = 3 to get y2=32−4⋅3+c=9−12+c=c−3 y_2 = 3^2 - 4 \cdot 3 + c = 9 - 12 + c = c - 3
step 4
Compare y1 y_1 and y2 y_2 to find y1=c y_1 = c and y2=c−3 y_2 = c - 3 , so y_1 > y_2 since c - 3 < c
Answer
y_1 > y_2
Key Concept
Substituting coordinates into the equation of a parabola to find specific values
Explanation
By substituting the x-coordinates of the given points into the parabola's equation, we can compare the y-values directly.
Solution by Steps
step 1
Given that the parabola y=x2−4x+c y = x^2 - 4x + c passes through (1,0) (1,0) , substitute x=1 x = 1 and y=0 y = 0 to find c c
step 2
Substituting gives 0=12−4⋅1+c 0 = 1^2 - 4 \cdot 1 + c , which simplifies to c=4−1=3 c = 4 - 1 = 3
step 3
Now that we know c c , the equation of the parabola is y=x2−4x+3 y = x^2 - 4x + 3
step 4
To find the other x-intercept, set y=0 y = 0 and solve 0=x2−4x+3 0 = x^2 - 4x + 3
step 5
Factoring the quadratic gives 0=(x−1)(x−3) 0 = (x - 1)(x - 3) . Since one root is x=1 x = 1 , the other root is x=3 x = 3
step 6
The coordinates of the other x-intercept are (3,0) (3,0)
Answer
The other x-intercept is at (3,0) (3,0) .
Key Concept
Finding x-intercepts of a parabola by setting y=0 y = 0 and solving the resulting quadratic equation
Explanation
The x-intercepts of a parabola are the points where the graph crosses the x-axis, which occur when y=0 y = 0 .
Solution by Steps
step 1
Given points A(x0,m) A(x_0, m) and B(x0+3,n) B(x_0 + 3, n) on the parabola y=x2−4x+3 y = x^2 - 4x + 3 with mn < 0 , we need to find the range of x0 x_0
step 2
Since mn < 0 , one of m m or n n is positive and the other is negative, which means one point is above and the other is below the x-axis
step 3
The parabola opens upwards and its vertex is at x=2 x = 2 , the midpoint between the x-intercepts x=1 x = 1 and x=3 x = 3
step 4
For A A to be below the x-axis and B B above, x0 x_0 must be less than the vertex x=2 x = 2
step 5
For B B to be below the x-axis and A A above, x0 x_0 must be greater than 1 1 and x0+3 x_0 + 3 must be less than 3 3 , so x_0 < 0
step 6
Combining the conditions, x0 x_0 must be in the range 1 < x_0 < 2
Answer
The range for x0 x_0 is 1 < x_0 < 2 .
Key Concept
Determining the range of x-values for points on a parabola where the product of their y-values is negative
Explanation
The product of the y-values of two points on a parabola being negative implies that one point lies above the x-axis and the other below, which restricts the x-values to a certain range.
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