Question

The vertex of a quadratic function is located at (1, 4), and the y-intercept of the function is (0, 1). What is the value of a if the function is written in the form y = a(x – h)2 + k?

Answer 1

Answer:

The value of a is $-3$

The equation is $y=-3(x-1)^{2}+4$

Step-by-step explanation:

we know that

The function of a vertical parabola written in vertex form is equal to

$y=a(x-h)^{2}+k$

where

(h,k) is the vertex of the parabola

in this problem we have

$(h,k)=(1,4)$

substitute in the equation

$y=a(x-1)^{2}+4$

Find the value of the coefficient a

Substitute the value of x and y of the point $(0,1)$ in the equation

$1=a(0-1)^{2}+4$

$1=a(-1)^{2}+4$

$1=a+4$

$a=-3$

The equation is equal to

$y=-3(x-1)^{2}+4$ ----> is a vertical parabola open downward

Answer 2

Vertex is at (1,4)  so h = 1 and  k = 4

so we have  y = a(x - 1)^2 + 4

when x = 0 y = 1 so

1 = a(0-1)^2 + 4

1 =  a + 4

a = -3