# How do you find the vertex, the y-intercept, and symmetric point, and use these to sketch the graph given y=3x^2-24x+17?

Nov 22, 2016

#### Explanation:

Let us convert the equation in to vertex form of equation of parabola $y = a {\left(x - h\right)}^{2} + k$. Note that in this form, while vertex is $\left(h , k\right)$, symmetric axis is $x = h$ and $y$-intercept can be obtained by putting $x = 0$.

Now, $y = 3 {x}^{2} - 24 x + 17$ can be written as

$y = 3 \left({x}^{2} - 8 x + 16\right) - 31$

or $y = 3 {\left(x - 4\right)}^{2} - 31$

Hence, vertex is $\left(4 , - 31\right)$ and as at $x = 0$, $y = 17$, $y$-intercept is $17$

and symmetric axis is $x = 4$.

Also intercept on $x$-axis are when $y = 0$, $x = 4 \pm \sqrt{\frac{31}{3}}$ i.e. $x = 7.21$ and$x = 0.79$.

Hence, curve looks like
graph{3x^2-24x+17=y[-5, 10, -50, 50]}