1. Find the discriminant of each of the following quadratic functions. Also find the number of x -intercepts of each of the functions.
(a) y=3x2−4x+3(b) y=2x2−4x−3(c) y=12x2+x−4(d) y=−x2+6x−9(e) y=−3x2−12x−7(f) y=−12x2−3x−4
(a) y=3x2−4x+3
Show/Hide Solution
y=3x2−4x+3 Comparing with ax2+bx+c we have, a=3, b=−4, c=3 ∴ Discriminant =b2−4ac =(−4)2−4(3)(3) =−20<0 ∴ There is no x−intercept point.
(b) y=2x2−4x−3
Show/Hide Solution
y=2x2−4x−3 Comparing with ax2+bx+c we have, a=2, b=−4, c=−3 ∴ Discriminant =b2−4ac =(−4)2−4(2)(−3) =8>0 ∴ There are two x−intercept points.
(c) y=12x2+x−4
Show/Hide Solution
y=12x2+x−4 Comparing with ax2+bx+c we have, a=12, b=1, c=−4 ∴ Discriminant =b2−4ac =(1)2−4(12)(−4) =9>0 ∴ There are two x−intercept points.
(d) y=−x2+6x−9
Show/Hide Solution
y=−x2+6x−9 Comparing with ax2+bx+c we have, a=−1, b=6, c=−9 ∴ Discriminant =b2−4ac =(6)2−4(−1)(−9) =0 ∴ There is one x−intercept point.
(e) y=−3x2−12x−7
Show/Hide Solution
y=−3x2−12x−7 Comparing with ax2+bx+c we have, a=−3, b=−12, c=−7 ∴ Discriminant =b2−4ac =(−12)2−4(−3)(−7) =60>0 ∴ There are two x−intercept points.
(f) y=−12x2−3x−4
Show/Hide Solution
y=−12x2−3x−4 Comparing with ax2+bx+c we have, a=−12, b=−3, c=−4 ∴ Discriminant =b2−4ac =(−3)2−4(−frac12)(−4) =1>0 ∴ There are two x−intercept points.
2. Find the intercept form of each of the quadratic functions. Also find the y -intercept, axis of symmetry, vertex, and range of each of the functions.
(a) y=2x2−2x−12(b) y=3x2−6x+3(c) y=12x2+x−4(d) y=2x2−5x−3(e) y=−6x2−7x+5(f) y=−12x2−3x−4
(a) y=2x2−2x−12
Show/Hide Solution
y=2x2−2x−12 y=2(x2−x−6) y=2(x+2)(x−3) When x=0, y=−12 ∴ y−intercept point=(0, −12) Comparing with y=a(x−p)(x−q), a=2, p=−2, q=3 ∴Axis of symmetry : x=p+q2⇒x=12 When x=12, y=2(12)2−2(12)−12=−252 ∴vertex =(12,−252) ∴range ={y | y≥−252}
(b) y=3x2−6x+3
Show/Hide Solution
y=3x2−6x+3 y=3(x2−2x+1) y=3(x−1)(x−1) When x=0, y=3 ∴ y−intercept point =(0, 3) Comparing with y=a(x−p)(x−q), a=3, p=1, q=1 ∴Axis of symmetry : x=1+12⇒x=1 When x=1, y=3(1−1)(1−1)=0 ∴vertex =(1,0)
(c) y=12x2+x−4
Show/Hide Solution
y=12x2+x−4 y=12(x2+2x−8) y=12(x+4)(x−2) When x=0, y=−4 ∴ y−intercept point =(0, −4) Comparing with y=a(x−p)(x−q), a=12, p=−4, q=2 ∴Axis of symmetry : x=−4+22⇒x=−1 When x=−1, y=12(−1+4)(−1−2)=−92 ∴vertex =(−1,−92) ∴range ={y | y≥−92}
(d) y=2x2−5x−3
Show/Hide Solution
y=2x2−5x−3 y=2(x2−52x−32) y=2(x+12)(x−3) When x=0, y=−3 ∴ y−intercept point =(0, −3) Comparing with y=a(x−p)(x−q), a=2, p=−12, q=3 ∴Axis of symmetry : x=−12+32⇒x=54 When x=54, y=2(54+12)(54−3)=−498 ∴vertex =(54,−498) ∴range ={y | y≥−498}
(e) y=−6x2−7x+5
Show/Hide Solution
y=−6x2−7x+5 y=−(6x2−7x−5) y=−(3x+5)(2x−1) y=−3(x+53)2(x−12) y=−6(x+53)(x−12) When x=0, y=5 ∴ y−intercept point =(0, 5) Comparing with y=a(x−p)(x−q), a=−6, p=−52, q=12 ∴Axis of symmetry : x=−53+122⇒x=−712 When x=54, y=−6(−712+53)(−712−12)=16924 ∴vertex =(−712,16924) ∴range ={y | y≤16924}
(f) y=−12x2−3x−4
Show/Hide Solution
y=−12x2−3x−4 y=−12(x2+6x+8) y=−12(x+4)(x+2) When x=0, y=−4 ∴ y−intercept point =(0, −4) Comparing with y=a(x−p)(x−q), a=−12, p=−4, q=−2 ∴Axis of symmetry : x=−4−22⇒x=−3 When x=54, y=−12(−3+4)(−3+2)=12 ∴vertex =(−3,12) ∴range ={y | y≤12}
စာဖတ်သူ၏ အမြင်ကို လေးစားစွာစောင့်မျှော်လျက်!