What is the vertex of $y = - 4 x^{2} - 7 x + 1$ $y= −4x^2 -7x + 1$ ?

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Answer 1 · 2017-05-23T04:08:18

From vertex form, the vertex is at $(- \frac{7}{8}, \frac{65}{16})$ $(-7/8 , 65/16)$ , which can be written as $(- .875, 4.0625)$ $(-.875, 4.0625)$

$y = - 4 x^{2} - 7 x + 1$ $y = -4x^2-7x+1$

Factor out the $- 4$ $-4$
$y = - 4 [x^{2} + \frac{7}{4} x - \frac{1}{4}]$ $y= -4 [x^2 + 7/4x -1/4]$

$y = - 4 [{(x + \frac{7}{8} x)}^{2} - \frac{49}{64} - \frac{1}{4}]$ $y = -4 [(x+7/8x)^2-49/64 - 1/4]$

$y = - 4 [{(x + \frac{7}{8} x)}^{2} - \frac{49 + 16}{64}]$ $y = -4 [(x+7/8x)^2-(49 + 16)/64]$

$y = - 4 [{(x + \frac{7}{8})}^{2} - \frac{65}{64}]$ $y = -4 [(x+7/8)^2 - 65/64]$

$y = - 4 {(x + \frac{7}{8})}^{2} + \frac{65}{16}$ $y = -4(x+ 7/8)^2 + 65/16$

From vertex form, the vertex is at $(- \frac{7}{8}, \frac{65}{16})$ $(-7/8 , 65/16)$ , which can be written as $(- .875, 4.0625)$ $(-.875, 4.0625)$

What is the vertex of y=−4x2−7x+1y= −4x^2 -7x + 1 ?