Determine the quadratic function whose vertex is (1, −5) and whose y-intercept is −3. (CRITICAL THINKING)?

Question

Precalc

Answer 1

More than one way to skin this cat....

The y intercept occurs when x = 0. so for this form of the equation:

y= ax^2 + bx + c

we have

-3 = a(0)^2 + b(0) + c 

or

-3 = c

so we know c.

We also know that is goes throw (1,-5) and so

-5 = a(1)^2 + b(1) -3

or  a + b = -2

The x location of the vertex is the average of the roots.  This makes the

vertex location  x = -b/(2a)

So we have  1 = -b/(2a)  which we can switch around to get

b = -2a

This can turn a + b = -2 into

a - 2a = -2 or

-a = -2

so a =2

and since a + b = -2 then 2 + b = -2, and b = -4

y = 2x^2 - 4x - 3

Answer 2

Vertex form of a quadratic is:

y = a(x - h)² + k

where the vertex is (h, k).  we are told this is (1, -5) so we can substitute those values:

y = a(x - 1)² - 5

The y-intercept is the value of y when x = 0.  So we can substitute those two values and can now solve for the remaining unknown:

-3 = a(0 - 1)² - 5

-3 = a(-1)² - 5

-3 = a - 5

2 = a

Now our equation is:

y = 2(x - 1)² - 5

Finally, expand to get your quadratic in standard form:

y = 2(x² - 2x + 1) - 5

y = 2x² - 4x + 2 - 5

y = 2x² - 4x - 3

Answer 3

Hint: (h,k) is the vertex, canonical form. Un-anon yourself and I will detail much more.

Hopefully no one will spoil you the answer thereby depriving you from your personal enhancement; that would be very inconsiderate of them.