Write the quadratic function \( f(x)=4 x^{2}-32 x+48 \) in intercept form

To express the quadratic function \( f(x)=4 x^{2}-32 x+48 \) in intercept form, we first need to factor it. Start by factoring out the common coefficient from the first two terms: \[ f(x) = 4(x^{2} - 8x + 12) \] Next, we apply the quadratic formula to find the x-intercepts (roots) of the equation \( x^{2} - 8x + 12 = 0 \): The solutions are given by: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] Here, \( a=1 \), \( b=-8 \), and \( c=12 \): \[ x = \frac{8 \pm \sqrt{(-8)^2 - 4 \cdot 1 \cdot 12}}{2 \cdot 1} \] \[ x = \frac{8 \pm \sqrt{64 - 48}}{2} \] \[ x = \frac{8 \pm \sqrt{16}}{2} \] \[ x = \frac{8 \pm 4}{2} \] This gives us the two roots: 1. \( x = \frac{12}{2} = 6 \) 2. \( x = \frac{4}{2} = 2 \) Now we can write the quadratic in intercept form as: \[ f(x) = 4(x - 6)(x - 2) \] And that’s the intercept form of your function!