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question 18 (mandatory) (1 point) for what value(s) of ( n ) does the f…

Question

question 18 (mandatory) (1 point)
for what value(s) of ( n ) does the function ( f(x) = 16x^2 - 8x + n ) have exactly one zero?

a) ( n = 1 )

b) ( n < 1 )

c) ( -1 < n < 1 )

d) ( n > -1 )

Explanation:

Step1: Recall discriminant formula

For a quadratic function \( f(x) = ax^2 + bx + c \), the discriminant is \( D = b^2 - 4ac \). A quadratic has exactly one zero when \( D = 0 \).
Here, \( a = 16 \), \( b = -8 \), \( c = n \).

Step2: Set discriminant to zero

Substitute into the discriminant formula: \( (-8)^2 - 4(16)(n) = 0 \).
Simplify: \( 64 - 64n = 0 \).

Step3: Solve for n

Subtract 64 from both sides: \( -64n = -64 \).
Divide both sides by -64: \( n = 1 \).

Answer:

a) \( n = 1 \)