Suppose one x-intercept of a parabola...

Question

kj4uts · Answer

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kj4uts · Answer

@Operand

anonymous · Answer

Some what, though I cannot remember if you use simple form (y = ax^2 +bx +c) or standard form to solve such a question.

When using standard form, you usually input everything, and then set the equation to zero, and factor to find the zeros (x-intercepts).

kj4uts · Answer

so this doesn't go in vertex form.

anonymous · Answer

$y = ax^2 + bx + c$ is simple form.
$y = a(x - h)^2 + k$ is standard form.

$y = a(x - (4.7))^2 + (9.2)$
Complete the square, to get the two zeros.

anonymous · Answer

You can also make 3 equations with 3 variables. You have the general form:
$$y=ax^2+bx+c$$
And then you have the vertex equation:
$$(T_x,T_y)=(-b/2a,-d/4a)$$
And then you can make the equations:
$$0=a*2^2+b*2+c$$$$4.7=-b/2*a$$$$9.2=-(b^2-4*a*c)/4*a$$
Now if you solve this you get the function, and then you can set it equal to 0 and find the other x-intercept.