# Finding features of parabolic formula

I have a question that requires me to find the intercepts and the turning point of a parabola from it's formula, but I'm not really sure how to find all this information out from the equation below. I tried using completing the square but I ended up with a weird answer that I don't believe is right. The equation is: $$y=-3x^2-5x+2$$ Thanks!!

The intercepts with the $X$-axis are the points of the form $(x,0)$. Take $y=0$ in the equation and solve it. You should obtain two points.

The intercepts with the $Y$-axis are the points of the form $(0,y)$. Take $x=0$ in the equation and solve it. You should obtain one point.

The turning point is the one where the first derivative becomes $0$. Thus, compute $y'$ in terms of $x$ and then look for the value of $x$ that makes the equation $0$.

Another form is to complete the square. You should obtain something of the form $y=a(x-b)^2+c$; then, the turning point is at $(b,y(b))$.

AugSB gave a perfect answer (+1!) but the shortcut all my students want to know is this:

The $x$-coordinate of the turning point (often called the vertex) is $x=-\frac{b}{2a}$. Plug this value into the equation to find the $y$-coordinate.

There are two intercepts. The $y$-intercept, and the $x$-intercepts.

$y$-intercept:

To find the $y$-intercept, you simply plug in $x=0$ into the equation to get the $y$ value. So thus, we have$$y=-3\cdot 0^2-5\cdot 0+2=2$$ so the $y$-intercept is $(0,2)$.

$x$-intercepts:

The $x$ intercepts are points that give $y=0$ when you plug in $y$. So plugging in $y=0$, we get $$-3x^2-5x+2=0\Leftrightarrow -(3x-1)(x+2)=0\implies x=\frac 13,x=-2$$ Therefore, the $x$-intercepts are $\left(\frac 13,0\right)$ and $(-2,0)$.

To find the Vertex, you can substitute $x$ with $\frac {-b}{2a}$ where given the quadratic of the form $ax^2+bx+c=0$.

In your example, the quadratic is $y=-3x^2-5x+2$ so we see that $a=-3,b=-5,c=2$. So we have $$-\frac b{2a}=-\frac 56\tag1$$ Substituting $x$ with $(1)$ gives the $y$ value:$$f\left(\frac 56\right)=-3\left(-\frac 56\right)^2+5\left(\frac 56\right)+2=\frac {49}{12}$$ And so the vertex lies on the point $\left(-\frac 56,\frac {49}{12}\right)$.

For parabola given by

$$a x^2 + b xc + c= 0$$

$y$ intercept is $y = c$ at $x = 0$

x-intercepts are when $y=0$

$$\frac{-b \pm \sqrt{ b^2-4 a c }}{2 a }$$

Vertex is average or middle point of the two above.

$$x= \frac{-b }{2 a }$$

This is also got differentiating and setting the obtained relation $2 ax+ b$ to zero.

Better to commit them to memory

Also better to remember max/min point of vertex

$$y= c- \frac{b^2}{4 a }$$.

EDIT:

When the above vanishes we have the two roots equal, or repeated roots. $y = a x^2 + b x + c$ is a perfect square. The parabola curve graph is tangent to $x$ axis at $x= - b/(2a)$.