1. **Graph the quadratic function** $y = x^2$. The function is a parabola opening upwards with vertex at the origin $(0,0)$. Given points: $$(-3,9), (-2,4), (-1,1), \left(-\frac{1}{2}, \frac{1}{4}\right), (0,0), \left(\frac{1}{2}, \frac{1}{4}\right), (1,1), (2,4), (3,9)$$ These points satisfy $y = x^2$ because squaring each $x$ value gives the corresponding $y$ value. 2. **Find the vertex of the quadratic function** $y = x^2 + 4x + 6$. Rewrite the quadratic in vertex form by completing the square: $$y = x^2 + 4x + 6 = (x^2 + 4x + 4) + 6 - 4 = (x + 2)^2 + 2$$ The vertex form is $y = (x + 2)^2 + 2$, so the vertex is at $$(-2, 2)$$ 3. **Find the value of $x$ in the right triangle** with legs 21, $x$, and hypotenuse 29. By the Pythagorean theorem: $$x^2 + 21^2 = 29^2$$ Calculate squares: $$x^2 + 441 = 841$$ Subtract 441 from both sides: $$x^2 = 841 - 441 = 400$$ Take the positive square root (lengths are positive): $$x = \sqrt{400} = 20$$ **Final answers:** - Graph of $y = x^2$ passes through the given points. - Vertex of $y = x^2 + 4x + 6$ is $(-2, 2)$. - The unknown leg $x$ in the triangle is $20$.