# How do you find the vertex and intercepts for f(x)=x^2-3x+4?

Jun 29, 2017

$\text{see explanation}$

#### Explanation:

$\text{the equation of a parabola in standard form }$

• y=ax^2+bx+c" has the x-coordinate of the vertex"

${x}_{\textcolor{red}{\text{vertex}}} = - \frac{b}{2 a}$

$f \left(x\right) = {x}^{2} - 3 x + 4 \text{ is in standard form}$

$\text{with } a = 1 , b = - 3 , c = 4$

$\Rightarrow {x}_{\textcolor{red}{\text{vertex}}} = - \frac{- 3}{2} = \frac{3}{2}$

$\text{substitute this value into f(x) for y-coordinate}$

$\Rightarrow {y}_{\textcolor{red}{\text{vertex}}} = {\left(\frac{3}{2}\right)}^{2} - 3 \left(\frac{3}{2}\right) + 4 = \frac{7}{4}$

$\Rightarrow \textcolor{m a \ge n t a}{\text{vertex }} = \left(\frac{3}{2} , \frac{7}{4}\right)$

$\textcolor{b l u e}{\text{Intercepts}}$

• " let x = 0 for y-intercept"

• " let y = 0 for x-intercepts"

$x = 0 \to y = 4 \leftarrow \textcolor{red}{\text{ y-intercept}}$

$y = 0 \to {x}^{2} - 3 x + 4 = 0$

$\text{check the value of the "color(blue)"discriminant}$

$\Delta = {b}^{2} - 4 a c = {\left(- 3\right)}^{2} - 16 = - 7$

$\text{since " Delta<0" roots are not real}$

$\Rightarrow \text{ there are no x-intercepts}$
graph{x^2-3x+4 [-10, 10, -5, 5]}