What is the quadratic function given the vertex is (2,3) and the y-intercept is (0,-5) ?

Answer 1

-2(x-2)²+3

Answer 2

Given that a quadratic equation is set up as: ax²+bx+c=y

and that any vertex on a quadratic curve is equal to 0, we solve the equation as such:

1.) Derivative of the base quadratic formula is: 2ax+b=y', where y' is the slope value at a given (x,y) point.

2.) Since the vertex has a slope of 0, we set the equation in (1.) as 2ax+b=0

3.) The vertex being at (2,3) can set up the derivative equation as: 2*a*2+b=0 -> 4a+b=0

4.) The given information of the y-int helps us set up the base quadratic equation as: a*0^2+b*0+c=-5

a.) This give us c = -5.

5.) From 4a., any point along the quadratic formula we are solving for takes the form of: ax^2+bx-5=y

6.) Taking the known vertex point, equation in (5.) becomes: a*(2)^2+b*(2)-5=3 -> 4a+2b=8

7.) Solving for simultaneous equations, 4a+b=0 and 4a+2b=8, multiplying (-1) by the first equation and adding that to the second equation gets b=8, thus plugging b in for either equation in (7.) gets a=-2

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The resulting equation becomes: -2x^2+8x-5=y

Testing for vertex: -2*(2)^2+8*2-5 = -8 + 16 - 5 = -13 + 16 = 3, so (2,3) does exist.

Answer 3

f(x) = a(x - 2)² + 3

-5 = a(0 - 2)² + 3

-8 = 4a

-2 = a

f(x) = -2(x - 2)² + 3

Answer 4

We can start with the vertex form. y = a(x - h)² + k, where (h, k) is the vertex of the parabola. Plugging in the values given for the vertex, you get:

y = a(x - 2)² + 3 Next, you can plug in the y intercept and this will allow you to solve for the a value.

-5=a(0 - 2)² + 3 Simplify

-5=a(- 2)² + 3

-5=4a + 3 Subtract 3 from both sides

-5-3=4a + 3-3 Simplify

-8=4a Divide both sides by 4

a=-2

Now plug in a

y = -2(x - 2)² + 3