# How do you use the important points to sketch the graph of y = 4x^2 + 5x - 1 ?

Jul 25, 2018

Below

#### Explanation:

To graph a parabola, it is a good idea to find:

• vertex
• x-intercepts
• y-intercepts

X-intercepts
When $y = 0$,
$x = \frac{- 5 \pm \sqrt{25 - 4 \left(4\right) \left(- 1\right)}}{2 \times 4}$
$x = \frac{- 5 \pm \sqrt{41}}{8}$
$x = \frac{- 5 + \sqrt{41}}{8}$ and $x = \frac{- 5 - \sqrt{41}}{8}$

Y-intercepts
When $x = 0$, $y = - 1$

Vertex
$x = - \frac{b}{2 a}$ where the quadratic equation is in the form $a {x}^{2} + b x + c$
$x = \frac{- 5}{2 \times 4}$
$x = - \frac{5}{8}$
Sub $x = - \frac{5}{8}$ into $y = 4 {x}^{2} + 5 x - 1$ to get the y-coordinate
$y = - 2 \frac{9}{16}$
The vertex is $\left(- \frac{5}{8} , - 2 \frac{9}{16}\right)$

Plotting your vertex and intercepts, you should get the graph below

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