The x-intercepts of \( y=3 x^{2}-8 x+1 \) are 0.13 and 2.54 Select one: True False

To find the x-intercepts of the quadratic function \( y=3x^{2}-8x+1 \), we set \( y \) to 0 and solve the equation \( 3x^{2}-8x+1=0 \). Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 3 \), \( b = -8 \), and \( c = 1 \), we find: \[ x = \frac{8 \pm \sqrt{(-8)^2 - 4 \cdot 3 \cdot 1}}{2 \cdot 3} = \frac{8 \pm \sqrt{64 - 12}}{6} = \frac{8 \pm \sqrt{52}}{6} = \frac{8 \pm 2\sqrt{13}}{6} = \frac{4 \pm \sqrt{13}}{3} \] Approximating \( \sqrt{13} \) gives about 3.61, leading to x-intercepts of approximately \( 0.13 \) and \( 2.54 \). Thus, the statement is true! So the answer is: True.