Write the following quadratic function in vertex form.. State the x and y-intercepts and the range. 𝑦 = −4𝑥^2 + 20𝑥 − 17
To write the quadratic function in vertex form, we need to complete the square.
Let's start by factoring out a -4 from the equation:
y = -4(x^2 - 5x + 17/4)
Next, we need to find the term to complete the square. We take half of the coefficient of x and square it:
(-5/2)^2 = 25/4
Now we add and subtract this term inside the parentheses:
y = -4(x^2 - 5x + 17/4 - 25/4)
Now we can rewrite the function in vertex form by grouping the terms:
y = -4((x^2 - 5x + 25/4 - 25/4) + 17/4)
y = -4((x - 5/2)^2 - 25/4 + 17/4)
Simplifying further:
y = -4((x - 5/2)^2 - 8/4)
y = -4((x - 5/2)^2 - 2)
Now the quadratic function is in vertex form: y = -4(x - 5/2)^2 - 2
The x-intercept can be found by setting y = 0:
0 = -4(x - 5/2)^2 - 2
4(x - 5/2)^2 = -2
(x - 5/2)^2 = -2/4
(x - 5/2)^2 = -1/2
Since the square of a real number cannot be negative, this equation does not have any real roots or x-intercepts.
The y-intercept can be found by setting x = 0:
y = -4(0 - 5/2)^2 - 2
y = -4(-5/2)^2 - 2
y = -4(25/4) - 2
y = -25 - 2
y = -27
Therefore, the y-intercept is y = -27.
To find the range, we can look at the coefficient of the squared term (-4). Since this coefficient is negative, the parabola opens downwards. Therefore, the range is (-∞, -2].
In summary, the quadratic function in vertex form is y = -4(x - 5/2)^2 - 2. It does not have any x-intercepts, the y-intercept is -27, and the range is (-∞, -2].