1. **State the problem:** Find the x- and y-intercepts of the parabola given by the equation $$y = 4x^2 - 12x + 9$$. 2. **Recall the definitions:** - The **x-intercepts** occur where $$y = 0$$. - The **y-intercept** occurs where $$x = 0$$. 3. **Find the y-intercept:** Substitute $$x = 0$$ into the equation: $$y = 4(0)^2 - 12(0) + 9 = 9$$ So, the y-intercept is at the point $$(0, 9)$$. 4. **Find the x-intercepts:** Set $$y = 0$$ and solve for $$x$$: $$0 = 4x^2 - 12x + 9$$ This is a quadratic equation. Use the quadratic formula: $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ where $$a = 4$$, $$b = -12$$, and $$c = 9$$. Calculate the discriminant: $$\Delta = b^2 - 4ac = (-12)^2 - 4(4)(9) = 144 - 144 = 0$$ Since the discriminant is zero, there is one real repeated root: $$x = \frac{-(-12)}{2 \times 4} = \frac{12}{8} = \frac{3}{2}$$ So, the x-intercept is at the point $$\left(\frac{3}{2}, 0\right)$$. **Final answer:** - x-intercept: $$\left(\frac{3}{2}, 0\right)$$ - y-intercept: $$(0, 9)$$