I don't know how to do this...

Write an equation for a parabola with​ x-intercepts (-4,0) and (2,0) which passes through the point (1,-10).

Answer:

y=2x^2+4x−16

Step-by-step explanation:

Assume that the equation of the parabola is y=ax2+bx+c.

Since the parabola passes through the point (−4,0), then 0=16a−4b+c.

Since the parabola passes through the point (2,0), then 0=4a+2b+c.

Since the parabola passes through the point (1,−10), then −10=a+b+c. Thus, we have obtained the following system:

Solving it, we get that a=2, b=4, c=−16. Thus, the equation of the parabola is y=2x^2+4x−16.

The equation of a parabola is  y = -x² + x + 4 with x-intercepts at (-4,0) and (2,0) that passes through the point (1,-10).

A parabola can be represented by the equation:

y = ax² + bx + c

To find the equation of a parabola that has x-intercepts at (-4,0) and (2,0), we can set y equal to 0 and solve for x to find the x-intercepts:

0 = a(-4)² + b(-4) + c

0 = a(2)² + b(2) + c

Solving these equations for a and c gives us:

c = b(-4)

c = -2b

The parabola passes through the point (1,-10), we can substitute the values of x and y into the equation to solve for b:

-10 = a(1)² + b(1) + (-2b)

-10 = a + b - 2b

-10 = a - b

Substituting c = b(-4) into the equation gives us:

y = a x² + b x + b(-4)

Substituting a = -b and solving for b gives us:

y = (-b) x² + b x + b(-4)

y = -b(x² - x - 4)

Since the x-intercepts are at (-4,0) and (2,0), the coefficient of x², which is -b, must be positive. Therefore, we can set b = 1 to get the equation of the parabola:

y = -(1)(x² - x - 4)

y = -x² + x + 4

Thus, the equation of a parabola is y = -x² + x + 4.

Learn more about the parabola here:

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