1. **State the problem:** We need to graph the function $y=2(x-4)^2-7$. 2. **Identify the type of function:** This is a quadratic function in vertex form $y=a(x-h)^2+k$ where $(h,k)$ is the vertex. 3. **Find the vertex:** Here, $h=4$ and $k=-7$, so the vertex is at $(4,-7)$. 4. **Determine the direction and width of the parabola:** Since $a=2$ (positive), the parabola opens upwards and is narrower than the standard parabola $y=x^2$ because $|2|>1$. 5. **Find the y-intercept:** Set $x=0$: $$y=2(0-4)^2-7=2(16)-7=32-7=25$$ So the y-intercept is $(0,25)$. 6. **Find the x-intercepts:** Set $y=0$ and solve for $x$: $$0=2(x-4)^2-7$$ $$2(x-4)^2=7$$ $$\cancel{2}(x-4)^2=\cancel{2}\frac{7}{2}$$ $$(x-4)^2=\frac{7}{2}$$ $$x-4=\pm\sqrt{\frac{7}{2}}$$ $$x=4\pm\sqrt{\frac{7}{2}}$$ 7. **Summary:** The vertex is at $(4,-7)$, the parabola opens upward and is narrower than $y=x^2$, the y-intercept is $(0,25)$, and the x-intercepts are $4\pm\sqrt{\frac{7}{2}}$. This information can be used to sketch the graph accurately.