Y=3(x-2)(x+1)(x-4) find all the zeros of the function
1 answer:
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We know that
<span>Vertex (0, 0) and directrix y = 4
is a vertical parabola------> open down (directrix above the vertex)
the equation is of the form
(y-k)=4p(x-h)</span>²--------> y=4p*x²-------> vertex (0,0)
y=4
y=h-p-----> y=-p
p=-4
the equation of the parabola is
y=-16x²
the answer is
y=-16x²
see the attached figure
<span>Vertex (0, 0) and directrix y = 4
is a vertical parabola------> open down (directrix above the vertex)
the equation is of the form
(y-k)=4p(x-h)</span>²--------> y=4p*x²-------> vertex (0,0)
y=4
y=h-p-----> y=-p
p=-4
the equation of the parabola is
y=-16x²
the answer is
y=-16x²
see the attached figure
Answer:
Use the given slope and point to substitute into the point-slope formula y−y1=m(x−x1).
Slope-intercept form:
y=3x²+4x−25
Point-slope form:
y+5=(3x+10)⋅(x−2)
Step-by-step explanation:
