How do you write a quadratic equation with x - intercepts: 2,3 point: (4,2)?

Question

Eli Assouline · Accepted Answer

#y = x² -5x + 6#

The general form of this type of quadratic is:

#y = k(x - r_1)(x - r_2)#

where #k# is a scaling factor that allows you to force the equation to pass through any given point and #r_1# and #r_2# are the x intercepts of the curve. Substituting the given information into the general form:

#2 = k(4 - 2)(4 - 3)#

#2 = k(2)(1)#

#k = 1#

The equation is:

#y = (x - 2)(x - 3)#

#y = x² -5x + 6#

Jace Anderson · Answer

undefined To write a quadratic equation with given x-intercepts and a point, you can use the factored form of a quadratic equation:

$$ f(x) = a(x - r)(x - s) $$

where $ r $ and $ s $ are the x-intercepts, and $ (h, k) $ is the given point. Then, plug in the given values for the x-intercepts and the point into the equation, and solve for $ a $. Once you find $ a $, substitute it back into the equation to obtain the quadratic equation.

Given x-intercepts $ r = 2 $ and $ s = 3 $, and point $ (4, 2) $, the equation becomes:

$$ f(x) = a(x - 2)(x - 3) $$

Now, plug in the point $ (4, 2) $:

$$ 2 = a(4 - 2)(4 - 3) $$
$$ 2 = a(2)(1) $$
$$ 2 = 2a $$
$$ a = 1 $$

Now that we have found $ a = 1 $, substitute it back into the equation:

$$ f(x) = 1(x - 2)(x - 3) $$
$$ f(x) = (x - 2)(x - 3) $$
$$ f(x) = x^2 - 5x + 6 $$

So, the quadratic equation with x-intercepts 2 and 3, and passing through the point (4, 2) is $ f(x) = x^2 - 5x + 6 $.