# Question #e5f61

Dec 20, 2017

Plot the solutions at $x = 1 , 3$; plot the vertex at $\left(2 , 1\right)$; plot the y-intercept at $y = - 3$, and you've got the graph!
graph{-x^2+4x-3 [-10, 10, -5, 5]}

#### Explanation:

You can find the solutions with the quadratic formula: $\frac{- 4 \pm \sqrt{{4}^{2} - 4 \left(- 1 \cdot - 3\right)}}{2 \cdot - 1} = 2 \pm - 1 = 1 , 3$
The $y$-intercept is easy to find by putting $0$ in for $x$, so the $y$-intercept is at $\left(0 , - 3\right)$.

Finally, plot the vertex by solving for the $x$-coordinate with $x = - \frac{b}{2 a} = \frac{- 4}{2 \cdot - 1} = 2$. Substitute $2$ for $x$ in the original equation to get the $y$-coordinate of the vertex: $- {2}^{2} + 4 \left(2\right) - 3 = 1$ So the vertex is at $\left(2 , 1\right)$.

If you need more points, make a table of values using numbers near the x-coordinate of the vertex.
Hope that helps!