Question

Please help! what is an equation of a parabola with x-intercepts at (2,0) and (-7,0) and which passes through the point (1,32)?​

Answer 1

Answer:

$f(x)=-4(x-2)(x+7)$

Or, in standard form:

$f(x)=-4x^2-20x+56$

Step-by-step explanation:

We can use the factored form of a quadratic equation:

$f(x)=a(x-p)(x-q)$

Where a is the leading coefficient and p and q are the zeros of the quadratic.

We know that the x-intercepts are at (2, 0) and (-7, 0).

So, let's substitute 2 for p and -7 for q. This yields:

$f(x)=a(x-2)(x+7)$

Now, we need to determine a.

We know that it passes through the point (1, 32). In other words, if we substitute 1 for x, we should get 32 for f(x). Therefore:

$32=a(1-2)(1+7)$

We can now solve for a. First, compute:

$32=a(-1)(8)$

Multiply:

$32=-8a$

Divide both sides by -8:

$a=-4$

So, the value of a is -4.

Therefore, our entire equation is:

$f(x)=-4(x-2)(x+7)$

Notes:

We can expand this into standard form:

$f(x)=-4(x-2)(x+7) \\ f(x)=-4(x^2+5x-14) \\ f(x)=-4x^2-20x+56$