Question

Write the equation for the parabola that has x− intercepts (−2,0) and (1.2,0) and y− intercept (0,−4).

Answer 1

Answer: The answer is $15y+4=25(x-1.6)^2.$

Step-by-step explanation:  We are to find the equation of the parabola with x-intercepts (-2,0), (1.2,0) and y-intercept (0,-4).

Therefore, axis of symmetry is

$x=\dfrac{1.2+2}{2}\\\\\Rightarrow x=1.6.$

Let the equation of the parabola be $y=a(x-1.6)^2+k.$

Substituting the points (x,y)=(0,-4) and (1.2,0) in the above equation, we find that

$-4=a(0-1.6)^2+k\\\\\Rightarrow k=-2.56a-4$

and

$0=a(1.2-1.6)^2+k\\\\\Rightarrow k=-0.16a.$

Solving the above two equations, we get

$a=-\dfrac{5}{3}~~\textup{and}~~k=-\dfrac{4}{15}.$

So, putting the values of 'a' and 'k' in the equation of the parabola, we get

$y=-\dfrac{5}{3}(x-1.6)^2-\dfrac{4}{15}\\\\\Rightarrow 15y+4=-25(x-1.6)^2.$

Thus, the equation of the parabola is  $15y+4=25(x-1.6)^2.$