Question

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`?

Answer 1

Please see below

Explanation:

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

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

`y=3(x^2-8x+16)-31`

or `y=3(x-4)^2-31`

Hence, vertex is `(4,-31)` 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+-sqrt(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]}