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How do I write this is standard form: y= 2(t-1)^2+12

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2 Answers

y = 2(t - 1)2 + 12
 
y = 2*(t - 1)*(t - 1) + 12
 
y = 2*[t2 - 2t +1] + 12
 
y = 2t2 -4t + 2 + 12
 
y = 2t2 - 4t + 14
 
Standard form:  In this case a=2, b = -4 and c = 14
Hi Megan;
y= 2(t-1)2+12
 
 
Let's begin with... (t-1)2
(t-1)(t-1)
Let's FOIL...
FIRST...(t)(t)=t2
OUTER...(t)(-1)=-t
INNER...(-1)(t)=-t
LAST...(-1)(-1)=1
t2-2t+1
 
Let's multiply by 2...
2(t2-2t+1)
2t2-4t+2
 
Let's add 12...
2t2-4t+2+12
2t2-4t+14