# How do you find the vertex and the intercepts for y = 1/4x^2 + 1/ 2x - 3/4?

Aug 20, 2017

Vertex is at $\left(- 1 , - 1\right)$, x intercepts are at $\left(- 3 , 0\right) \mathmr{and} \left(1 , 0\right)$
and y intercept is at $\left(0 , - \frac{3}{4}\right)$,

#### Explanation:

$y = \frac{1}{4} {x}^{2} + \frac{1}{2} x - \frac{3}{4}$ . Comparing with standard equation

$y = a {x}^{2} + b + c$ we get $a = \frac{1}{4} , b = \frac{1}{2} , c = - \frac{3}{4}$ We know

vertex $\left(x\right) = - \frac{b}{2 a} = - \frac{1}{2} / \left(2 \cdot \frac{1}{4}\right) = - 1$. Putting $x = - 1$

in the equation $y = \frac{1}{4} {x}^{2} + \frac{1}{2} x - \frac{3}{4}$,

we get, vertex $\left(y\right) = \frac{1}{4} {\left(- 1\right)}^{2} + \frac{1}{2} \left(- 1\right) - \frac{3}{4} = \frac{1}{4} - \frac{1}{2} - \frac{3}{4}$

$= - \frac{4}{4} = - 1$ .Vertex is at $\left(- 1 , - 1\right)$ . Putting $x = 0$

in the equation we can get y intercept as $y = - \frac{3}{4} \mathmr{and} \left(0 , - \frac{3}{4}\right)$,

putting $y = 0$ in the equation we can get x intercepts as

$0 = \frac{1}{4} {x}^{2} + \frac{1}{2} x - \frac{3}{4} \mathmr{and} {x}^{2} + 2 x - 3 = 0$

or $\left(x + 3\right) \left(x - 1\right) = 0 \therefore x = - 3 \mathmr{and} x = 1$ , x intercepts are at

$\left(- 3 , 0\right) \mathmr{and} \left(1 , 0\right)$

graph{1/4x^2+1/2x-3/4 [-10, 10, -5, 5]}

Aug 20, 2017

Alternative approach
${y}_{\text{intercept}} \to \left(x , y\right) = \left(0 , - \frac{3}{4}\right)$

Vertex$\to \left(x , y\right) = \left(- 1 , - 1\right)$

${x}_{\text{intercepts}} \to x = - 3 \mathmr{and} x = 1$

#### Explanation:

Given: $y = \frac{1}{4} {x}^{2} + \frac{1}{2} x - \frac{3}{4}$

Compare to the form of $y = a {x}^{2} + b x + c$

$\textcolor{b l u e}{\text{Determine the y intercept}}$

${y}_{\text{intercept}} = c = - \frac{3}{4} \to \left(x , y\right) = \left(0 , - \frac{3}{4}\right)$
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

$\textcolor{b l u e}{\text{Determine the vertex}}$

You have three methods:

1. Using the completed square form equation you can read it directly off but with a small amount of adjustment.
Complete breakdown on this can be found on https://socratic.org/s/aHr8G5h4

2.factorising or use the formula to determine the $x$ intercepts and the $x$ value of the vertex will be half way between

1. You can do it this way:
Note that this process is part of that for completing the square.

Write as:

$y = \frac{1}{4} \left({x}^{2} \textcolor{red}{+ 2 x}\right) - \frac{3}{4}$

${x}_{\text{vertex}} = \left(- \frac{1}{2}\right) \times \left(\textcolor{red}{+ 2}\right) = - 1$

${y}_{\text{vertex}} = \frac{1}{4} {\left(- 1\right)}^{2} + \frac{1}{2} \left(- 1\right) - \frac{3}{4} = - 1$

Vertex$\to \left(x , y\right) = \left(- 1 , - 1\right)$
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$\textcolor{b l u e}{\text{Determine the "x_("interpts}}$

Using other method
$y = a {x}^{2} + b x + c = 0 \textcolor{w h i t e}{. .}$ where $x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

a=1/4; b=1/2; c= -3/4

$x = \frac{- \frac{1}{2} \pm \sqrt{{\left(\frac{1}{2}\right)}^{2} - 4 \left(\frac{1}{4}\right) \left(- \frac{3}{4}\right)}}{2 \times \frac{1}{4}}$

$x = - 1 \pm 2 \sqrt{\frac{1}{4} + \frac{3}{4}}$

$x = - 1 \pm 2$

$x = - 3 \mathmr{and} x = 1$