Find, in the form y=zx2+bz+c, the equation of the quadratic

Lloyd Allen

Lloyd Allen

Answered question

2021-12-02

Find, in the form y=zx2+bz+c, the equation of the quadratic function whose graph:
a) touches axis x at 4 and passes through the point (2,12)
b) has x-intercepts 3 and -2 and y-intercept 3.

Answer & Explanation

Huses1969

Huses1969

Beginner2021-12-03Added 18 answers

Soln y=ax2+6x+c
(2,12) is a point on the curve
12=4a+2b+c-(1)
touches axis z at 4
(4,0) is pt. on the curve
0=16a+4b+c --(2)
y=0 is atangest at (4,0)
(since it 'touches' axis )
THe slpoe of the tangest is given
by m=y=2zx+6
0=8a+6b=8a
put b=8a in eqn(2)
16a+4(8a)+c=0
16a32a+c=0
c=16a
Now put b=8a and c=16 in eqn(1)
4a+2(8a)+(16a)=12
4a16a+15a=12
4a=12
a=12/4=3
2. b=8a=24
and c=16a=48
Therefore, y=3x224x+48.

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