Question

question 18 (mandatory) (1 point) saved
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 by -64: \( n = 1 \).

Answer:

a) \( n = 1 \)