Write 1.81 as a fraction and an improper number
1 answer:
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We know that
The Standard Form of a Quadratic Equation<span> looks like this
</span> ax²<span> + bx + c = 0
</span>
we have
<span>x -1 0 1 2 3
y -20 -6 2 4 0
for x=0 y=-6
then
</span> y=ax² + bx + c --------> -6=a*0² + b*0 + c ---------> c=-6
for y=0 x=3
then
y=ax² + bx + c-------> 0=a*3² + b*3 -6---------> 9a+3b=6----> equation 1
The Standard Form of a Quadratic Equation<span> looks like this
</span> ax²<span> + bx + c = 0
</span>
we have
<span>x -1 0 1 2 3
y -20 -6 2 4 0
for x=0 y=-6
then
</span> y=ax² + bx + c --------> -6=a*0² + b*0 + c ---------> c=-6
for y=0 x=3
then
y=ax² + bx + c-------> 0=a*3² + b*3 -6---------> 9a+3b=6----> equation 1
for x=2 y=4
then
y=ax² + bx + c-----> 4=a*2² + b*2 -6-----> 4=4a+2b-6-----> 4a+2b=10----> a=2.5-0.5b----> equation 2
I substitute 2 in 1
9*[2.5-0.5b]+3b=6------> 22.5-4.5b+3b=6------> 1.5b=16.5------> b=11
a=2.5-0.5*11------> a=2.5-5.5------> a=-3
The Standard Form of a Quadratic Equation is
ax² + bx + c = 0--------> -3x²+11x-6=0
the answer is
-3x²+11x-6=0
See the attached figure
9514 1404 393
Answer:
58. 2 real roots; see attached for a graph
59. -6, 3
Step-by-step explanation:
58. A graphing calculator graphs this easily. The graph is shown in the attachment. There are 2 x-intercepts, hence 2 real roots.
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59. The x-intercepts are (-6, 0) and (3, 0), so the x-values x=-6 and x=3 satisfy the equation y = 0.
