# How do you graph the parabola y=x^2-3x+8 using vertex, intercepts and additional points?

Dec 3, 2017

For $y = a {x}^{2} + b x + c$ the $x$ for vertex is

${x}_{v} = - \frac{b}{2 a} = \frac{3}{2}$

so

y_v=x^2−3x+8 = (3/2)^2-3*3/2+8= 23/4

So the vertex is $\left(\frac{3}{2} , \frac{23}{4}\right)$.

Since $a > 0$, so it is looking up.
The $y$-intercept is $Y = {0}^{2} - 3 \cdot 0 + 8 = 8$ so $\left(0 , 8\right)$

#### Explanation:

After finding the $y$-intercept, we should practically find the $x$-intercept, but as you see, the graph is looking upward since $a$ is bigger than $0$ and the vertex is higher than $0$.

So there will be no $x$-intercept, meaning that the graph will not intersect the $x$-axis. You can just plug in any number and then use the symmetrical characteristic of the hyperbola and draw the whole thing.

Dec 3, 2017

Vertex: $\left(\setminus \frac{3}{2} , \setminus \frac{23}{4}\right)$

$y$-intercept: $\left(0 , 8\right)$

Other point: $\left(3 , 8\right)$

#### Explanation:

We can find the $x$-coordinate of the vertex by using the formula $- \setminus \frac{b}{2 a}$, where $a$ and $b$ come from the standard form of $a {x}^{2} + b x + c$.

In this equation,

• $a = 1$
• $b = - 3$
• $c = 8$

Plugging in yield:

$- \setminus \frac{- 3}{2 \left(1\right)} \setminus \quad \setminus \implies \setminus \quad \setminus \frac{3}{2}$

To find the $y$-coordinate of the vertex, plug the $x$-coordinate into the equation in standard form:

${x}^{2} - 3 x + 8$

$\setminus \implies {\left(\setminus \frac{3}{2}\right)}^{2} - 3 \left(\setminus \frac{3}{2}\right) + 8$

$\setminus \implies \setminus \frac{9}{4} - \setminus \frac{9}{2} + 8$

$\setminus \implies - \setminus \frac{9}{4} + 8$

$\setminus \implies \setminus \frac{23}{4}$

$\setminus \therefore$ the vertex is $\left(\setminus \frac{3}{2} , \setminus \frac{23}{4}\right)$.

The $y$-intercept is simply $\left(0 , c\right)$, which is $\left(0 , 8\right)$ in this case.

For a third point, we can find the axis of symmetry of the parabola, and reflect the $y$-intercept across that line.

The axis of symmetry is the vertex’s $x$-coordinate, which is $1.5$.

$\setminus \therefore$ the third point will have an $x$-coordinate of $1.5 \setminus \cdot 2$, which is $3$. The $y$-coordinate is the same as the $y -$-intercept, which is $8$.

$\setminus \therefore$ a third point is $\left(3 , 8\right)$.

That’s all we need to graph a parabola.