1. **Problem:** Find the x and y-intercepts of the quadratic function $f(x) = x^2 + 3x - 4$. 2. **Formula and rules:** - The y-intercept is found by evaluating $f(0)$. - The x-intercepts are found by solving $f(x) = 0$. - For a quadratic $ax^2 + bx + c = 0$, use the quadratic formula: $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ 3. **Find the y-intercept:** $$f(0) = 0^2 + 3(0) - 4 = -4$$ So, the y-intercept is $(0, -4)$. 4. **Find the x-intercepts:** Set $f(x) = 0$: $$x^2 + 3x - 4 = 0$$ Here, $a=1$, $b=3$, $c=-4$. Calculate the discriminant: $$\Delta = b^2 - 4ac = 3^2 - 4(1)(-4) = 9 + 16 = 25$$ Since $\Delta > 0$, there are two real roots. Apply the quadratic formula: $$x = \frac{-3 \pm \sqrt{25}}{2(1)} = \frac{-3 \pm 5}{2}$$ Calculate each root: - $$x_1 = \frac{-3 + 5}{2} = \frac{2}{2} = 1$$ - $$x_2 = \frac{-3 - 5}{2} = \frac{-8}{2} = -4$$ So, the x-intercepts are $(1, 0)$ and $(-4, 0)$. **Final answer:** - y-intercept: $(0, -4)$ - x-intercepts: $(1, 0)$ and $(-4, 0)$