Asked by John
f(x) = x^2 + 10x + 24
=(x^2 + 10x)+ 24
=(x^2 + 10x + 25 - 25) + 24
f(x) =(x+5)^2 - 1
Vertex: (-5, -1)
How do you sketch this? I know that in the graph one of the coord is (-5, -1) but how do I find the others in order to make a parabola? :|
=(x^2 + 10x)+ 24
=(x^2 + 10x + 25 - 25) + 24
f(x) =(x+5)^2 - 1
Vertex: (-5, -1)
How do you sketch this? I know that in the graph one of the coord is (-5, -1) but how do I find the others in order to make a parabola? :|
Answers
Answered by
MathMate
Factorize to get the zeroes:
f(x)=x²+10x+24=(x+6)(x+4)
Therefore the zeroes at at x=-6, x=-4.
The y-intercept is y=24 (= constant term)
The coefficient of x² is positive, so the curve is concave upwards.
The vertex is the minimum of the parabola, and x=-5 passing through the vertex is the axis of symmetry.
That should give enough information to sketch the graph?
f(x)=x²+10x+24=(x+6)(x+4)
Therefore the zeroes at at x=-6, x=-4.
The y-intercept is y=24 (= constant term)
The coefficient of x² is positive, so the curve is concave upwards.
The vertex is the minimum of the parabola, and x=-5 passing through the vertex is the axis of symmetry.
That should give enough information to sketch the graph?
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