What is the vertices x=1-y^2 ?

The vertices x=1-y^2 is (h,k)=(1,0),p=-1/4

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vertices x=1-y^2

vertices

x = 1 - y^{2}

(h, k) = (1, 0), p = - \frac{1}{4}

Solution steps

x = 1 - y^{2}

Rewrite

x = 1 - y^{2}

in the standard form:

4 (- \frac{1}{4}) (x - 1) = (y - 0)^{2}

Therefore parabola properties are:

(h, k) = (1, 0), p = - \frac{1}{4}

f (x) = 8 x - 13 - x^{2}

- \frac{( x + 3 ) ^{2}}{20} + 1

x^{2} - 2 x + 2

f (x) = - 0.01 x^{2} + 0.7 x + 6

y = + x^{2} - 3