The graph of f(x)=ax2+bx+c is shown below, such that b2−4ac=−4. If the length of segment AB and AC are 1 and 4 respectively, then the value of (a+b+c) is equal to
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Solution
f(x)=ax2+bx+c D=b2−4ac=−4...(1) Least value of f(x)=−D4a=¯¯¯¯¯¯¯¯AB ⇒−D4a=1 ⇒−(−4)=−4a[From(1)] ⇒a=1
Also −b2a=−4⇒b=8 And b2−4ac=−4 ⇒64−4c=−4 ⇒c=17 Hence, a+b+c=1+8+17=26