Question 231995
If the equation is in the form
{{{y = ax^2 + bx + c}}} , then the x component
of the vertex is at {{{-(b/2a)))
given:
vertex is at (6,-5), so
(1) {{{6 = -(b/2a)}}}
and, plugging the point into general equation,
{{{y = ax^2 + bx + c}}}
{{{-5 = a*6^2 + b*6 + c}}}
(2) {{{-5 = 36a + 6b + c}}}
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The y-intercept is at (0,175), so plugging this into
general equation:
{{{175 = a*0^2 + b*0 + c}}}
(3) {{{175 = c}}}
Now I have 3 equations and 3 unknowns, so it's solvable
From (1),
(1) {{{6 = -(b/2a)}}}
{{{b = -12a}}}
Plug this into (2)
(2) {{{-5 = 36a + 6b + c}}}
(2) {{{-5 = 36a + 6*(-12a) + c}}}
also, from(3),
(2) {{{-5 = 36a + 6*(-12a) + 175}}}
{{{36a - 72a = -175 -5}}}
{{{-36a = -180}}}
{{{a = 5}}}
and, since
{{{b = -12a}}}
{{{b = -12*5}}}
{{{b = -60}}}
Now I can write the actual equation
{{{y = 5x^2 - 60x + 175}}}
Set {{{y = 0}}} to find x-intercepts
Then divide through by {{{5}}}
{{{x^2 - 12x + 35 = 0}}}
I can see right away that it factors into
{{{(x - 5)(x - 7) = 0}}}
{{{x = 5}}} and {{{x = 7}}} are the solutions
So, the x-intercepts are at (5,0) and (7,0)