Asked by Anonymous
How?? I only know how to convert standard form functions into intercept form, but I do not understand how to do this.
Convert the functions into standard form.
1.y=2(x-4)(x+3)
2.y-2=-3(x+4)^2
Convert the functions into standard form.
1.y=2(x-4)(x+3)
2.y-2=-3(x+4)^2
Answers
Answered by
Reiny
Both of your equations are quadratics
the standard form, sometimes called the vertex form, of a quadratic takes the form
y = a(x-h)^2 + k , where the vertex is (h,k)
your first equation is in intercept form, so the x of the vertex must be midway between -3 and 4 which would be 1/2
sub in 1/2 for x to get y
y = 2(1/2-4)(1/2+3) = -49/2
so y = 2(x - 1/2)^2 - 49/2
another way would be to first expand it
y = 2(x^2 - x - 12) = 2x^2 - 2x - 24
now completing the square ....
y = 2(x^2 - x + 1/4 -1/4) - 24
= 2( x - 1/2_)^2 - 1/2 - 24
= 2(x-1/2)^2 - 49/2 , same as above
for your second, we are almost there, just add 2 to both sides
y = -3(x+4)^2 + 2
the standard form, sometimes called the vertex form, of a quadratic takes the form
y = a(x-h)^2 + k , where the vertex is (h,k)
your first equation is in intercept form, so the x of the vertex must be midway between -3 and 4 which would be 1/2
sub in 1/2 for x to get y
y = 2(1/2-4)(1/2+3) = -49/2
so y = 2(x - 1/2)^2 - 49/2
another way would be to first expand it
y = 2(x^2 - x - 12) = 2x^2 - 2x - 24
now completing the square ....
y = 2(x^2 - x + 1/4 -1/4) - 24
= 2( x - 1/2_)^2 - 1/2 - 24
= 2(x-1/2)^2 - 49/2 , same as above
for your second, we are almost there, just add 2 to both sides
y = -3(x+4)^2 + 2
Answered by
Anonymous
Oh okay thank you, it's very clear now!
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