1. **State the problem:** Find the points where the quadratic graph $y = 5x^2 + 23x + 12$ crosses the $x$-axis and $y$-axis. 2. **Recall the rules:** - The graph crosses the $x$-axis where $y=0$. So solve $5x^2 + 23x + 12 = 0$. - The graph crosses the $y$-axis where $x=0$. So find $y$ when $x=0$. 3. **Find the $y$-intercept:** Substitute $x=0$ into the equation: $$y = 5(0)^2 + 23(0) + 12 = 12$$ So the $y$-intercept is at $(0, 12)$. 4. **Find the $x$-intercepts:** Solve the quadratic equation: $$5x^2 + 23x + 12 = 0$$ Use the quadratic formula: $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ where $a=5$, $b=23$, $c=12$. Calculate the discriminant: $$\Delta = b^2 - 4ac = 23^2 - 4 \times 5 \times 12 = 529 - 240 = 289$$ Calculate the roots: $$x = \frac{-23 \pm \sqrt{289}}{2 \times 5} = \frac{-23 \pm 17}{10}$$ Two solutions: - $$x_1 = \frac{-23 + 17}{10} = \frac{-6}{10} = -0.6$$ - $$x_2 = \frac{-23 - 17}{10} = \frac{-40}{10} = -4$$ 5. **Write the $x$-intercepts as points:** - $(-0.6, 0)$ - $(-4, 0)$ **Final answers:** - $x$-intercepts: $(-0.6, 0)$ and $(-4, 0)$ - $y$-intercept: $(0, 12)$