How do you graph the parabola $y = x^2 + 4x + 1$ using vertex, intercepts and additional points?

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Answer 1 · 2018-07-20T05:21:06

Vertex $(-2, -3)$
Y-intercept $(0,1)$
x-intercepts $(0.268, 0); (-3.732, 0)$

Given -

$y=x^2+4x+1$

Vertex

$x=(-b)/(2a)=(-4)/2=-2$

At $x=-2; y=(-2)^2+4(-2)+1=4-8+1=-3$

Vertex $(-2, -3)$

Y-intercept
At $x=0; y=(0)^2+4(0)+1=0+0+1=1$

Y-intercept $(0,1)$

x_intercept
At $y=0; x^2+4x-1=0$

$x^2+4x=-1$

$x^2+4x+4=-1+4=3$

$(x+2)^2=3$

$x+2=+-sqrt3=+-1.732$

$x=1.732-2=0.268$

$x=-1.732-2=-3.732$

x-intercepts $(0.268, 0); (-3.732, 0)$

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How do you graph the parabola y = x^2 + 4x + 1 y = x^2 + 4x + 1 using vertex, intercepts and additional points?