题目 #16486

(1)因为点P(−1, a)在直线\(l_{2}\): y = 2x + 4上，∴\(2 \times (−1) + 4 =\)a, 即 a = 2, 则P的坐标为(−1,2), 设直线\(l_{1}\)的解析式为：y = kx + h(k ≠0)，那么{k + b = 0−k + b = 2, 解得：{k = −1b = 1∴\(l_{1}\)的解析式为：y = −x + 1； 1 (2)∵直线\(l_{2}\)与x轴相交于点A，∴A点的坐标为(−2,0), 则AB = 3，∴S_{\(\Delta{}ABP\)}= _{2}\(\times 3 \times 2 = 3\). 2.(1)a = −4; (2) −11 < y < 5. (1)∵一次函数y = ax + 1的图象经过点(1, −3), ∴a + 1 = −3, ∴a = −4. (2)∵一次函数的函数表达式为y = −4x + 1, ∴当x = −1时, y = 5; 当x = 3时, y=−11, ∵a < 0, ∴ y随x的增大而减小, ∴当−1 < x < 3时, y的取值范围为−11 < y < 5． 3.(1) −2; (2)2; (3)\(y_{1}\)>\(y_{2}\)>\(y_{3}\). (1)将点(2, −4)代入正比例函数y = kx, 可得−4 = 2k, 解得k = −2； (2)由(1)可知, 该正比例函数解析式为y = −2x, 将点(−1, m)代入, 可得m=−\(2 \times (−1) = 2\)； (3)对于正比例函数y = −2x, ∵k = −2 < 0, ∴函数y随x的增大而减小,又∵\(x_{1}\)<\(x_{2}\)<\(x_{3}\), ∴\(y_{1}\)>\(y_{2}\)>\(y_{3}\). 4.(1)y = −x + 4; (2) {x = 2 y = 2 ; (3)3． (1)把点C(m, 2)代入y = 2x −2，得2m −2 = 2，解得m = 2，∴ 点 C(2,2), 把点B(3,1), C(2,2)代入y = kx + b，得{3k + b = 12k + b = 2，解得{k = −1b = 4, ∴直线\(l_{2}\)的表达式为y = −x + 4； (2)∵直线y = 2x −2与y = −x + 4的交点C的坐标为(2,2)，∴{y = 2x −2y = kx + b的解为{x = 2y = 2; (3)在y = 2x −2中，令y = 0，得2x −2 = 0，解得x = 1，∴ 点 D(1,0), 在y = −x + 4中，令y = 1 1 0，得−x + 4 = 0，解得x = 4，∴点A(4,0)，∴AD = 3, ∴S_{\(\Delta{}ADC\)}= _{2}AD ⋅|\(y_{C}\)| = _{2}\(\times 3 \times 2 = 3\). 5.(1)y = 2x −1; (2)点 C(−3, −7)在这条直线上; (3)x > _{2}.^{1} (1)设这个一次函数的表达式为y = kx + b(k ≠0), 把A(3,5)和B(−4, −9)代入到y = kx + b(k ≠0)中得 {3k + b = 5−4k + b = −9 , 解得{k = 2b = −1 , ∴这个一次函数的表达式为y = 2x −1； (2)在y = 2x −1中, 当x = −3时, y =\(2 \times (−3)\)−1 = −7, ∴该直线经过点(−3, −7),所以点C(−3, −7) 在这条直线上； (3)在 y = 2x −1 中,当 y = 2x −1 = 0 时, x =\(1 _{2}\), ∵一次项系数为2, 是正数, ∴y随x增大而增大, ∴当y > 0 时, x >\(1_{2}\)． 6.(1)y = −3x; (2)点A在图象上; (3)\(x_{1}\)>\(x_{2}\)时,\(y_{1}\)<\(y_{2}\). (1)将点(2, −6)代入y = kx, 得2k = −6, 解得k = −3, ∴这个函数解析式为y = −3x． (2)当x = −1时, y = (−3)\(\times (−1) = 3\)∴点A(−1,3)在这个函数图象上． (3)∵k = −3 < 0, ∴y随着x增大而减小, ∵图象上的两点C(\(x_{1}\),\(y_{1}\)), D(\(x_{2}\),\(y_{2}\)),且\(x_{1}\)>\(x_{2}\), ∴\(y_{1}\)<\(y_{2}\). 7.(1)C(2,4); (2)y = 2x; (3)20. (1)把C(m, 4)代入 y = − 1\(1 _{2}\)x + 5, 得：4 = − _{2}m + 5, 解得m = 2；∴C(2,4); (2)设直线\(l_{2}\)的解析式为y = kx, 把C(2,4)代入, 得：4 = 2k, 解得k = 2，∴直线\(l_{2}\)的解析式为y = 2x； (3)∵y = − 1\(1 _{2}\)x + 5，∴当y = − _{2}x + 5 = 0时, 解得x = 10，所以A(10,0) ∴OA = 10, ∵C(2,4), ∴S_{\(\Delta{}AOC\)}= 1\(1 _{2}\)OA ⋅\(y_{C}\)= _{2}\(\times 10 \times 4 = 20\). 8.(1)m = 2; (2)3． (1)∵函数y = (m + 2)\(x^{m}^{2}^{−3}\)−1是一次函数, ∴\(m^{2}\)−3 = 1, 解得m = ±2, ∵m + 2 ≠0,所以m = 2; (2)将m = 2代入得一次函数解析式为y = 4x −1, ∴y随x的增大而增大, ∴当−3 ≤x ≤1时, 当x = 1时 , y有最大值, 最大值为y =\(4 \times 1\)−1 = 3． 9.(1)a = 1, b = 2; (2)y = −x + 2; (3)\(y_{1}\)<\(y_{2}\). (1)由题意, 将点B(−a, 3)代入y = −3x得: 3a = 3, 解得a = 1, 将点A(0,2)代入y=kx + b得: b = 2． (2)由(1)可知, y = kx + 2, 且B(−1,3), 将点B(−1,3)代入y = kx + 2得: −k + 2 = 3,解得k = −1, 则一次函数y = kx + b的解析式为y = −x + 2． (3)在一次函数y = −x + 2中, y随x的增大而减小, ∵P(m,\(y_{1}\)), Q(m −1,\(y_{2}\))是一次函数y = −x + 2图象上的两点, 且m > m −1, ∴\(y_{1}\)<\(y_{2}\). 10.y = −2x + 3． 因为一次函数y = kx + b，当−3 ≤x ≤1时，1 ≤y ≤9，且y随着x的增大而减小，∴ 当 x = −3 时 , y = 9; 当 x = 1 时, y = 1, ∴{−3k + b = 9, 解得{k = −2 k + b = 1 b = 3,所以一次函数的解析式为：y = −2x + 3． 11.(1)y = − 3\(7 _{2}\).^\({1}_{3} _{2}\)x − _{2}; (2) − (1)∵y + 2与x + 1成正比例, 则设y + 2 = k(x + 1)(k ≠0), 当x = 1时, y = −5即−5+2 = k(1 + 1), 解得 k = − 3 3\(7 _{2}\), 即y + 2 = − _{2}(x + 1), 整理得y = − _{2}x − _{2}； (2)由 y = − 3\(7 _{2}\)x − _{2}可知− _{2}< 0, ∴y随x的增大而减小, ∴当−2 ≤x ≤3时函数的最大值为\(y_{max}\)= 3 7 1 − _{2}\(\times (−2)\)− _{2}= − _{2}． 12.(1)正比例函数的解析式是y = −2x; (2)\(y_{1}\)>\(y_{2}\). (1)设正比例函数的解析式是y = kx(k ≠0), ∵当x = 3时, y = −6, ∴3k = −6,解得k = −2, ∴ 正比例函数的解析式是y = −2x； (2)∵−2 < 0, ∴y随x的增大而减小, 又a < a + 1, ∴\(y_{1}\)>\(y_{2}\). 13.(1)\(y_{1}\)= 200 + 5x;\(y_{2}\)= 216 + 4.5x；(2)方案①更省钱，付款350元. (1)方案①:\(y_{1}\)=\(30 \times 8 + 5(x −8) = 200 + 5x\)方案②:\(y_{2}\)= (\(30 \times 8 + 5x\))\(\times 90% = 216 + 4.5x\)； (2)由题意可得:\(y_{1}\)= 200 + 5x = 200 + 150 = 350(元),\(y_{2}\)= 216 + 4.5x = 351 (元), 故方案①更省钱, 付款350元. 14.(1)得\(y_{1}\)= 10x + 400,\(y_{2}\)= 15x, b的实际意义是一张羽毛球健身的年卡的费用为400元； (2)方案一费用少些，见解析 (1)根据题意，得直线\(y_{1}\)=\(k_{1}\)x + b经过点(0,400), (120,1600), ∴{1600 =\(120k^{1}\)+ b,解得{k1= 10 b = 400 b = 400 , ∴ 直线的解析式为\(y_{1}\)= 10x + 400，b的实际意义是一张羽毛球健身的年卡的费用为400元；由题意得\(y_{2}\)= 15x. (2)根据题意，两种方案费用相等的次数满足方程10x + 400 = 15x，解得x = 80，当x > 80时，\(y_{2}\)> 365\(y_{1}\)；当x < 80时，\(y_{1}\)>\(y_{2}\)；∵每周去俱乐部打球2次(365天)，∴一年打球次数至少为x ≥ _{7}\(\times 2\)≥ 104 > 80，故\(y_{2}\)>\(y_{1}\)，所以选择方案一费用少些． 15.(1)m = 3; y = _{3}x + 2; (2) {x = 32y = 4 ; (3)0 < x < 3．